Quiz 1: Quantitative Chemistry

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1. Calcium carbonate decomposes on heating as shown below. «math xmlns=¨https://www.w3.org/1998/Math/MathML¨»«mi»C«/mi»«mi»a«/mi»«mi»C«/mi»«msub»«mi»O«/mi»«mn»3«/mn»«/msub»«mo»§nbsp;«/mo»«mo»§#8594;«/mo»«mo»§nbsp;«/mo»«mi»C«/mi»«mi»a«/mi»«mi»O«/mi»«mo»§nbsp;«/mo»«mo»+«/mo»«mo»§nbsp;«/mo»«mi»C«/mi»«msub»«mi»O«/mi»«mn»2«/mn»«/msub»«/math» When 50 g of calcium carbonate are decomposed, 7 g of calcium oxide are formed. What is the percentage yield of calcium oxide?

Explanation

«math xmlns=¨http://www.w3.org/1998/Math/MathML¨»«msub»«mi»M«/mi»«mi»r«/mi»«/msub»«mo»(«/mo»«mi»C«/mi»«mi»a«/mi»«mi»C«/mi»«msub»«mi»O«/mi»«mn»3«/mn»«/msub»«mo»)«/mo»«mo»§nbsp;«/mo»«mo»=«/mo»«mo»§nbsp;«/mo»«mn»100«/mn»«mo»§nbsp;«/mo»«mi»g«/mi»«mo»§nbsp;«/mo»«mi»m«/mi»«mi»o«/mi»«msup»«mi»l«/mi»«mrow»«mo»-«/mo»«mn»1«/mn»«/mrow»«/msup»«/math»

Amount of «math xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mi»C«/mi»«mi»a«/mi»«mi»C«/mi»«msub»«mi»O«/mi»«mn»3«/mn»«/msub»«mo»§nbsp;«/mo»«mo»=«/mo»«mo»§nbsp;«/mo»«mfrac»«mrow»«mn»50«/mn»«mo»§nbsp;«/mo»«mi»g«/mi»«/mrow»«mrow»«mn»100«/mn»«mo»§nbsp;«/mo»«mi»g«/mi»«/mrow»«/mfrac»«mo»§nbsp;«/mo»«mo»=«/mo»«mo»§nbsp;«/mo»«mn»0«/mn»«mo».«/mo»«mn»50«/mn»«mo»§nbsp;«/mo»«mi»m«/mi»«mi»o«/mi»«mi»l«/mi»«/math»

Stoichiometric ratio of «math xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mi»C«/mi»«mi»a«/mi»«mi»C«/mi»«msub»«mi»O«/mi»«mn»3«/mn»«/msub»«mo»§nbsp;«/mo»«mo»:«/mo»«mo»§nbsp;«/mo»«mi»C«/mi»«mi»a«/mi»«mi»O«/mi»«/math» is 1 : 1

Amount of CaO = Amount of «math xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mi»C«/mi»«mi»a«/mi»«mi»C«/mi»«msub»«mi»O«/mi»«mn»3«/mn»«/msub»«/math» = 0.50 mol

«math xmlns=¨http://www.w3.org/1998/Math/MathML¨»«msub»«mi»M«/mi»«mi»r«/mi»«/msub»«mo»(«/mo»«mi»C«/mi»«mi»a«/mi»«mi»O«/mi»«mo»)«/mo»«mo»§nbsp;«/mo»«mo»=«/mo»«mo»§nbsp;«/mo»«mn»56«/mn»«mo»§nbsp;«/mo»«mi»g«/mi»«/math»

Mass of CaO = 0.50 x 56 = 28 g

% yield = 7 g/28 g x 100% = 25%
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Quiz 1: Quantitative Chemistry - Quiz

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2. When the equation below is balanced for 1 mol of «math xmlns=¨https://www.w3.org/1998/Math/MathML¨»«msub»«mi»C«/mi»«mn»3«/mn»«/msub»«msub»«mi»H«/mi»«mn»4«/mn»«/msub»«/math», what is the coefficient for «math xmlns=¨https://www.w3.org/1998/Math/MathML¨»«msub»«mi»O«/mi»«mn»2«/mn»«/msub»«/math»? «math xmlns=¨https://www.w3.org/1998/Math/MathML¨»«msub»«mi»C«/mi»«mn»3«/mn»«/msub»«msub»«mi»H«/mi»«mn»4«/mn»«/msub»«mo»§nbsp;«/mo»«mo»+«/mo»«mo»§nbsp;«/mo»«mi»_«/mi»«mi»_«/mi»«msub»«mi»O«/mi»«mn»2«/mn»«/msub»«mo»§nbsp;«/mo»«mo»§#8594;«/mo»«mo»§nbsp;«/mo»«mi»_«/mi»«mi»_«/mi»«mi»C«/mi»«msub»«mi»O«/mi»«mn»2«/mn»«/msub»«mo»§nbsp;«/mo»«mo»+«/mo»«mo»§nbsp;«/mo»«mi»_«/mi»«mi»_«/mi»«msub»«mi»H«/mi»«mn»2«/mn»«/msub»«mi»O«/mi»«/math»

Explanation

«math xmlns=¨http://www.w3.org/1998/Math/MathML¨»«msub»«mi»C«/mi»«mn»3«/mn»«/msub»«msub»«mi»H«/mi»«mn»4«/mn»«/msub»«mo»§nbsp;«/mo»«mo»+«/mo»«mo»§nbsp;«/mo»«mn»4«/mn»«msub»«mi»O«/mi»«mn»2«/mn»«/msub»«mo»§nbsp;«/mo»«mo»§#8594;«/mo»«mo»§nbsp;«/mo»«mn»3«/mn»«mi»C«/mi»«msub»«mi»O«/mi»«mn»2«/mn»«/msub»«mo»§nbsp;«/mo»«mo»+«/mo»«mo»§nbsp;«/mo»«mn»2«/mn»«msub»«mi»H«/mi»«mn»2«/mn»«/msub»«mi»O«/mi»«/math»
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3. What is the number of oxygen atoms in one mole of «math xmlns=¨https://www.w3.org/1998/Math/MathML¨»«mi»C«/mi»«mi»u«/mi»«mi»S«/mi»«msub»«mi»O«/mi»«mn»4«/mn»«/msub»«mo».«/mo»«mn»5«/mn»«msub»«mi»H«/mi»«mn»2«/mn»«/msub»«mi»O«/mi»«/math»?

Explanation

1 mol of «math xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mi»C«/mi»«mi»u«/mi»«mi»S«/mi»«msub»«mi»O«/mi»«mn»4«/mn»«/msub»«mo».«/mo»«mn»5«/mn»«msub»«mi»H«/mi»«mn»2«/mn»«/msub»«mi»O«/mi»«/math» = 9 mol of O atoms

Number of O atoms = «math xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mn»9«/mn»«mo»§nbsp;«/mo»«mi»x«/mi»«mo»§nbsp;«/mo»«mn»6«/mn»«mo».«/mo»«mn»0«/mn»«mo»§nbsp;«/mo»«mi»x«/mi»«mo»§nbsp;«/mo»«msup»«mn»10«/mn»«mn»23«/mn»«/msup»«mo»§nbsp;«/mo»«mo»=«/mo»«mo»§nbsp;«/mo»«mn»54«/mn»«mo»§nbsp;«/mo»«mi»x«/mi»«mo»§nbsp;«/mo»«msup»«mn»10«/mn»«mn»23«/mn»«/msup»«mo»§nbsp;«/mo»«mo»=«/mo»«mo»§nbsp;«/mo»«mn»5«/mn»«mo».«/mo»«mn»4«/mn»«mo»§nbsp;«/mo»«mi»x«/mi»«mo»§nbsp;«/mo»«msup»«mn»10«/mn»«mn»24«/mn»«/msup»«mo»§nbsp;«/mo»«mi»a«/mi»«mi»t«/mi»«mi»o«/mi»«mi»m«/mi»«mi»s«/mi»«/math»

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4.«math xmlns=¨https://www.w3.org/1998/Math/MathML¨»«mi»d«/mi»«msup»«mi»m«/mi»«mn»3«/mn»«/msup»«/math» of carbon monoxide, CO(g), and 2 «math xmlns=¨https://www.w3.org/1998/Math/MathML¨»«mi»d«/mi»«msup»«mi»m«/mi»«mn»3«/mn»«/msup»«/math» of oxygen, «math xmlns=¨https://www.w3.org/1998/Math/MathML¨»«msub»«mi»O«/mi»«mn»2«/mn»«/msub»«/math»(g), at the same temperature and pressure are mixed together. Assuming complete reaction according to the equation given, what is the maximum volume of carbon dioxide, «math xmlns=¨https://www.w3.org/1998/Math/MathML¨»«mi»C«/mi»«msub»«mi»O«/mi»«mn»2«/mn»«/msub»«/math»(g), in «math xmlns=¨https://www.w3.org/1998/Math/MathML¨»«mi»d«/mi»«msup»«mi»m«/mi»«mn»3«/mn»«/msup»«/math», that can be formed? «math xmlns=¨https://www.w3.org/1998/Math/MathML¨»«mn»2«/mn»«mi»C«/mi»«mi»O«/mi»«mo»(«/mo»«mi»g«/mi»«mo»)«/mo»«mo»§nbsp;«/mo»«mo»+«/mo»«mo»§nbsp;«/mo»«msub»«mi»O«/mi»«mn»2«/mn»«/msub»«mo»(«/mo»«mi»g«/mi»«mo»)«/mo»«mo»§nbsp;«/mo»«mo»§#8594;«/mo»«mo»§nbsp;«/mo»«mn»2«/mn»«mi»C«/mi»«msub»«mi»O«/mi»«mn»2«/mn»«/msub»«mo»(«/mo»«mi»g«/mi»«mo»)«/mo»«/math»

Explanation

Apply the Avogadro's law: V proportional to n

Stoichiometric ratio of «math xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mi»C«/mi»«mi»O«/mi»«mo»§nbsp;«/mo»«mo»:«/mo»«mo»§nbsp;«/mo»«mi»C«/mi»«msub»«mi»O«/mi»«mn»2«/mn»«/msub»«/math» is 2 : 2 (or 1 : 1)

Expected volume of «math xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mi»C«/mi»«msub»«mi»O«/mi»«mn»2«/mn»«/msub»«/math» produced from CO = 5 «math xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mi»d«/mi»«msup»«mi»m«/mi»«mn»3«/mn»«/msup»«/math»

Stoichiometric ratio of «math xmlns=¨http://www.w3.org/1998/Math/MathML¨»«msub»«mi»O«/mi»«mn»2«/mn»«/msub»«mo»§nbsp;«/mo»«mo»:«/mo»«mo»§nbsp;«/mo»«mi»C«/mi»«msub»«mi»O«/mi»«mn»2«/mn»«/msub»«/math» is 1 : 2

Expected volume of «math xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mi»C«/mi»«msub»«mi»O«/mi»«mn»2«/mn»«/msub»«/math» produced from «math xmlns=¨http://www.w3.org/1998/Math/MathML¨»«msub»«mi»O«/mi»«mn»2«/mn»«/msub»«/math» = 2 x 2 = 4 «math xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mi»d«/mi»«msup»«mi»m«/mi»«mn»3«/mn»«/msup»«/math»

SInce «math xmlns=¨http://www.w3.org/1998/Math/MathML¨»«msub»«mi»O«/mi»«mn»2«/mn»«/msub»«/math» gives smaller volume of «math xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mi»C«/mi»«msub»«mi»O«/mi»«mn»2«/mn»«/msub»«/math» produced, therefore «math xmlns=¨http://www.w3.org/1998/Math/MathML¨»«msub»«mi»O«/mi»«mn»2«/mn»«/msub»«/math» is limiting reactant and the maximum volume of «math xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mi»C«/mi»«msub»«mi»O«/mi»«mn»2«/mn»«/msub»«/math» produced is «math xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mi mathvariant=¨bold-italic¨»d«/mi»«msup»«mi mathvariant=¨bold-italic¨»m«/mi»«mn»3«/mn»«/msup»«/math»

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5. Which sample of nitrogen gas, «math xmlns=¨https://www.w3.org/1998/Math/MathML¨»«msub»«mi»N«/mi»«mn»2«/mn»«/msub»«/math», contains the greatest number of nitrogen molecules?

Explanation

Number of molecules is proportional to amount

Amount of «math xmlns=¨http://www.w3.org/1998/Math/MathML¨»«msub»«mi»N«/mi»«mn»2«/mn»«/msub»«/math» in choice 1 = 1.4/28 = 1/20 mol

Amount of «math xmlns=¨http://www.w3.org/1998/Math/MathML¨»«msub»«mi»N«/mi»«mn»2«/mn»«/msub»«/math» in choice 2 = 1.4/22.4 = 14/224 = 1/16 mol

Amount of «math xmlns=¨http://www.w3.org/1998/Math/MathML¨»«msub»«mi»N«/mi»«mn»2«/mn»«/msub»«/math» in choice 3 = «math xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mn»1«/mn»«mo».«/mo»«mn»4«/mn»«mo»§nbsp;«/mo»«mi»x«/mi»«mo»§nbsp;«/mo»«msup»«mn»10«/mn»«mn»23«/mn»«/msup»«mo»/«/mo»«mn»6«/mn»«mo».«/mo»«mn»0«/mn»«mo»§nbsp;«/mo»«mi»x«/mi»«mo»§nbsp;«/mo»«msup»«mn»10«/mn»«mn»23«/mn»«/msup»«mo»§nbsp;«/mo»«mo»=«/mo»«mo»§nbsp;«/mo»«mn»14«/mn»«mo»/«/mo»«mn»60«/mn»«mo»§nbsp;«/mo»«mo»=«/mo»«mo»§nbsp;«/mo»«mn»7«/mn»«mo»/«/mo»«mn»30«/mn»«mo»§nbsp;«/mo»«mi»m«/mi»«mi»o«/mi»«mi»l«/mi»«/math»

Amount of «math xmlns=¨http://www.w3.org/1998/Math/MathML¨»«msub»«mi»N«/mi»«mn»2«/mn»«/msub»«/math» in choice 4 = 1.4 mol

Hence choice 4 has the greatest amount of «math xmlns=¨http://www.w3.org/1998/Math/MathML¨»«msub»«mi»N«/mi»«mn»2«/mn»«/msub»«/math» and therefore the greatest number of «math xmlns=¨http://www.w3.org/1998/Math/MathML¨»«msub»«mi»N«/mi»«mn»2«/mn»«/msub»«/math» molecules.


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6. Which compound has the empirical formula with the largest mass?

Explanation

Empirical formula of «math xmlns=¨http://www.w3.org/1998/Math/MathML¨»«msub»«mi»C«/mi»«mn»2«/mn»«/msub»«msub»«mi»H«/mi»«mn»6«/mn»«/msub»«mo»§nbsp;«/mo»«mo»=«/mo»«mo»§nbsp;«/mo»«mi»C«/mi»«msub»«mi»H«/mi»«mn»3«/mn»«/msub»«/math»

Empirical formula of «math xmlns=¨http://www.w3.org/1998/Math/MathML¨»«msub»«mi»C«/mi»«mn»2«/mn»«/msub»«msub»«mi»H«/mi»«mn»4«/mn»«/msub»«mo»§nbsp;«/mo»«mo»=«/mo»«mo»§nbsp;«/mo»«mi»C«/mi»«msub»«mi»H«/mi»«mn»2«/mn»«/msub»«/math»

Empirical formula of «math xmlns=¨http://www.w3.org/1998/Math/MathML¨»«msub»«mi»C«/mi»«mn»2«/mn»«/msub»«msub»«mi»H«/mi»«mn»2«/mn»«/msub»«mo»§nbsp;«/mo»«mo»=«/mo»«mo»§nbsp;«/mo»«mi»C«/mi»«mi»H«/mi»«/math»

Empirical formula of «math xmlns=¨http://www.w3.org/1998/Math/MathML¨»«msub»«mi»C«/mi»«mn»3«/mn»«/msub»«msub»«mi»H«/mi»«mn»6«/mn»«/msub»«mo»§nbsp;«/mo»«mo»=«/mo»«mo»§nbsp;«/mo»«mi»C«/mi»«msub»«mi»H«/mi»«mn»2«/mn»«/msub»«/math»

Hence the empirical formula with the largest mass is «math xmlns=¨http://www.w3.org/1998/Math/MathML¨»«msub»«mi»C«/mi»«mn»2«/mn»«/msub»«msub»«mi»H«/mi»«mn»6«/mn»«/msub»«/math»

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7. A gas sample occupies a volume «math xmlns=¨https://www.w3.org/1998/Math/MathML¨»«msub»«mi»V«/mi»«mn»1«/mn»«/msub»«/math» at a pressure «math xmlns=¨https://www.w3.org/1998/Math/MathML¨»«msub»«mi»P«/mi»«mn»1«/mn»«/msub»«/math» and a Kelvin temperature «math xmlns=¨https://www.w3.org/1998/Math/MathML¨»«msub»«mi»T«/mi»«mn»1«/mn»«/msub»«/math». What would be the temperature of the gas, «math xmlns=¨https://www.w3.org/1998/Math/MathML¨»«msub»«mi»T«/mi»«mn»2«/mn»«/msub»«/math», if both its pressure and volume are doubled?

Explanation

Applying the equation «math xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mfrac»«mrow»«msub»«mi»P«/mi»«mn»1«/mn»«/msub»«msub»«mi»V«/mi»«mn»1«/mn»«/msub»«/mrow»«msub»«mi»T«/mi»«mn»1«/mn»«/msub»«/mfrac»«mo»§nbsp;«/mo»«mo»=«/mo»«mo»§nbsp;«/mo»«mfrac»«mrow»«msub»«mi»P«/mi»«mn»2«/mn»«/msub»«msub»«mi»V«/mi»«mn»2«/mn»«/msub»«/mrow»«msub»«mi»T«/mi»«mn»2«/mn»«/msub»«/mfrac»«/math»

Pressure double => «math xmlns=¨http://www.w3.org/1998/Math/MathML¨»«msub»«mi»P«/mi»«mn»2«/mn»«/msub»«mo»§nbsp;«/mo»«mo»=«/mo»«mo»§nbsp;«/mo»«mn»2«/mn»«msub»«mi»P«/mi»«mn»1«/mn»«/msub»«/math»

Volume double => «math xmlns=¨http://www.w3.org/1998/Math/MathML¨»«msub»«mi»V«/mi»«mn»2«/mn»«/msub»«mo»§nbsp;«/mo»«mo»=«/mo»«mo»§nbsp;«/mo»«mn»2«/mn»«msub»«mi»V«/mi»«mn»1«/mn»«/msub»«/math»

«math xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mfrac»«mrow»«msub»«mi»P«/mi»«mn»1«/mn»«/msub»«msub»«mi»V«/mi»«mn»1«/mn»«/msub»«/mrow»«msub»«mi»T«/mi»«mn»1«/mn»«/msub»«/mfrac»«mo»§nbsp;«/mo»«mo»=«/mo»«mo»§nbsp;«/mo»«mfrac»«mrow»«mo»(«/mo»«mn»2«/mn»«msub»«mi»P«/mi»«mn»1«/mn»«/msub»«mo»)«/mo»«mo»(«/mo»«mn»2«/mn»«msub»«mi»V«/mi»«mn»1«/mn»«/msub»«mo»)«/mo»«/mrow»«msub»«mi»T«/mi»«mn»2«/mn»«/msub»«/mfrac»«mo»§nbsp;«/mo»«mo»§#8658;«/mo»«mo»§nbsp;«/mo»«mfrac»«mn»1«/mn»«msub»«mi»T«/mi»«mn»1«/mn»«/msub»«/mfrac»«mo»§nbsp;«/mo»«mo»=«/mo»«mo»§nbsp;«/mo»«mfrac»«mn»4«/mn»«msub»«mi»T«/mi»«mn»2«/mn»«/msub»«/mfrac»«mo»§nbsp;«/mo»«mo»§#8658;«/mo»«mo»§nbsp;«/mo»«msub»«mi mathvariant=¨bold-italic¨»T«/mi»«mn»2«/mn»«/msub»«mo»§nbsp;«/mo»«mo»=«/mo»«mo»§nbsp;«/mo»«mn»4«/mn»«msub»«mi mathvariant=¨bold-italic¨»T«/mi»«mn»1«/mn»«/msub»«/math»

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8. What volume, in «math xmlns=¨https://www.w3.org/1998/Math/MathML¨»«mi»c«/mi»«msup»«mi»m«/mi»«mn»3«/mn»«/msup»«/math», of 0.200 «math xmlns=¨https://www.w3.org/1998/Math/MathML¨»«mi»m«/mi»«mi»o«/mi»«mi»l«/mi»«mo»§nbsp;«/mo»«mi»d«/mi»«msup»«mi»m«/mi»«mrow»«mo»-«/mo»«mn»3«/mn»«/mrow»«/msup»«/math» HCl(aq) is required to neutralize 25.0 «math xmlns=¨https://www.w3.org/1998/Math/MathML¨»«mi»c«/mi»«msup»«mi»m«/mi»«mn»3«/mn»«/msup»«/math» of 0.200 «math xmlns=¨https://www.w3.org/1998/Math/MathML¨»«mi»m«/mi»«mi»o«/mi»«mi»l«/mi»«mo»§nbsp;«/mo»«mi»d«/mi»«msup»«mi»m«/mi»«mrow»«mo»-«/mo»«mn»3«/mn»«/mrow»«/msup»«/math» «math xmlns=¨https://www.w3.org/1998/Math/MathML¨»«mi»B«/mi»«mi»a«/mi»«mo»(«/mo»«mi»O«/mi»«mi»H«/mi»«msub»«mo»)«/mo»«mn»2«/mn»«/msub»«/math»(aq)?

Explanation

Assume the given volume is interms of «math xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mi»d«/mi»«msup»«mi»m«/mi»«mn»3«/mn»«/msup»«/math»

«math xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mn»2«/mn»«mi»H«/mi»«mi»C«/mi»«mi»l«/mi»«mo»§nbsp;«/mo»«mo»+«/mo»«mo»§nbsp;«/mo»«mi»B«/mi»«mi»a«/mi»«mo»(«/mo»«mi»O«/mi»«mi»H«/mi»«msub»«mo»)«/mo»«mn»2«/mn»«/msub»«mo»§nbsp;«/mo»«mo»§#8594;«/mo»«mo»§nbsp;«/mo»«mi»B«/mi»«mi»a«/mi»«mi»C«/mi»«msub»«mi»l«/mi»«mn»2«/mn»«/msub»«mo»§nbsp;«/mo»«mo»+«/mo»«mo»§nbsp;«/mo»«mn»2«/mn»«msub»«mi»H«/mi»«mn»2«/mn»«/msub»«mi»O«/mi»«/math»

Amount of «math xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mi»B«/mi»«mi»a«/mi»«mo»(«/mo»«mi»O«/mi»«mi»H«/mi»«msub»«mo»)«/mo»«mn»2«/mn»«/msub»«/math» = 25.0 x 0.200 = 5.00 mol

Stoichiometric ratio of «math xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mi»H«/mi»«mi»C«/mi»«mi»l«/mi»«mo»§nbsp;«/mo»«mo»:«/mo»«mo»§nbsp;«/mo»«mi»B«/mi»«mi»a«/mi»«mo»(«/mo»«mi»O«/mi»«mi»H«/mi»«msub»«mo»)«/mo»«mn»2«/mn»«/msub»«/math» is 2 : 1

Amount of HCl = 5.00 x 2 = 10.0 mol

Concentration of HCl = 0.200 «math xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mi»m«/mi»«mi»o«/mi»«mi»l«/mi»«mo»§nbsp;«/mo»«mi»d«/mi»«msup»«mi»m«/mi»«mrow»«mo»-«/mo»«mn»3«/mn»«/mrow»«/msup»«/math»

Volume of HCl required = 10.0/0.200 = 50.0 «math xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mi mathvariant=¨bold-italic¨»c«/mi»«msup»«mi mathvariant=¨bold-italic¨»m«/mi»«mn»3«/mn»«/msup»«/math»

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9. A blast furnace contains 1600 kg of iron(III) oxide («math xmlns=¨https://www.w3.org/1998/Math/MathML¨»«msub»«mi»M«/mi»«mi»r«/mi»«/msub»«/math» = 160) and 144 kg of carbon («math xmlns=¨https://www.w3.org/1998/Math/MathML¨»«msub»«mi»A«/mi»«mi»r«/mi»«/msub»«/math» = 12). Assuming that they react according to the following equation: «math xmlns=¨https://www.w3.org/1998/Math/MathML¨»«mi»F«/mi»«msub»«mi»e«/mi»«mn»2«/mn»«/msub»«msub»«mi»O«/mi»«mn»3«/mn»«/msub»«mo»(«/mo»«mi»s«/mi»«mo»)«/mo»«mo»§nbsp;«/mo»«mo»+«/mo»«mo»§nbsp;«/mo»«mn»3«/mn»«mi»C«/mi»«mo»(«/mo»«mi»s«/mi»«mo»)«/mo»«mo»§nbsp;«/mo»«mo»§#8594;«/mo»«mo»§nbsp;«/mo»«mn»2«/mn»«mi»F«/mi»«mi»e«/mi»«mo»(«/mo»«mi»s«/mi»«mo»)«/mo»«mo»§nbsp;«/mo»«mo»+«/mo»«mo»§nbsp;«/mo»«mn»3«/mn»«mi»C«/mi»«mi»O«/mi»«mo»(«/mo»«mi»g«/mi»«mo»)«/mo»«/math» what is the limiting reagent and the maximum theoretical yield of iron?

Explanation

Assume the mass is given in grams

Stoichiometric ratio of «math xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mi»F«/mi»«msub»«mi»e«/mi»«mn»2«/mn»«/msub»«msub»«mi»O«/mi»«mn»3«/mn»«/msub»«/math» is 1 : 2

Amount of «math xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mi»F«/mi»«msub»«mi»e«/mi»«mn»2«/mn»«/msub»«msub»«mi»O«/mi»«mn»3«/mn»«/msub»«/math» = 1600/160 = 10 mol

Expected amount of Fe formed from «math xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mi»F«/mi»«msub»«mi»e«/mi»«mn»2«/mn»«/msub»«msub»«mi»O«/mi»«mn»3«/mn»«/msub»«/math» = 2 x 10 = 20 mol

Stoichiometric ratio of C : Fe is 3 : 2

Amount of C = 144/12 = 12 mol

Expected amount of Fe fromed from C = 12 x (2/3) = 8 mol

Since carbon gives the least amount of Fe, therefore carbon is the limiting reactant

«math xmlns=¨http://www.w3.org/1998/Math/MathML¨»«msub»«mi»M«/mi»«mi»r«/mi»«/msub»«mo»(«/mo»«mi»F«/mi»«mi»e«/mi»«mo»)«/mo»«mo»§nbsp;«/mo»«mo»=«/mo»«mo»§nbsp;«/mo»«mn»56«/mn»«mo»§nbsp;«/mo»«mi»g«/mi»«mo»§nbsp;«/mo»«mi»m«/mi»«mi»o«/mi»«msup»«mi»l«/mi»«mrow»«mo»-«/mo»«mn»1«/mn»«/mrow»«/msup»«/math»

Maximum yield of iron = 8 x 56 = 448 kg

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10. What will be the concentration of sulfate ions in «math xmlns=¨https://www.w3.org/1998/Math/MathML¨»«mi»m«/mi»«mi»o«/mi»«mi»l«/mi»«mo»§nbsp;«/mo»«mi»d«/mi»«msup»«mi»m«/mi»«mrow»«mo»-«/mo»«mn»3«/mn»«/mrow»«/msup»«/math» when 0.20 mol «math xmlns=¨https://www.w3.org/1998/Math/MathML¨»«mi»K«/mi»«mi»A«/mi»«mi»l«/mi»«mo»(«/mo»«mi»S«/mi»«msub»«mi»O«/mi»«mn»4«/mn»«/msub»«msub»«mo»)«/mo»«mn»2«/mn»«/msub»«/math» is dissolved in water to give 100 «math xmlns=¨https://www.w3.org/1998/Math/MathML¨»«mi»c«/mi»«msup»«mi»m«/mi»«mn»3«/mn»«/msup»«/math» of aqueous solution?

Explanation

1 mol of «math xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mi»K«/mi»«mi»A«/mi»«mi»l«/mi»«mo»(«/mo»«mi»S«/mi»«msub»«mi»O«/mi»«mn»4«/mn»«/msub»«msub»«mo»)«/mo»«mn»2«/mn»«/msub»«/math» contains 2 mol of «math xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mi»S«/mi»«msup»«msub»«mi»O«/mi»«mn»4«/mn»«/msub»«mrow»«mn»2«/mn»«mo»-«/mo»«/mrow»«/msup»«/math» ions

0.20 mol  of «math xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mi»K«/mi»«mi»A«/mi»«mi»l«/mi»«mo»(«/mo»«mi»S«/mi»«msub»«mi»O«/mi»«mn»4«/mn»«/msub»«msub»«mo»)«/mo»«mn»2«/mn»«/msub»«/math» = 0.40 mol of «math xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mi»S«/mi»«msup»«msub»«mi»O«/mi»«mn»4«/mn»«/msub»«mrow»«mn»2«/mn»«mo»-«/mo»«/mrow»«/msup»«/math» ions

Volume = 100 «math xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mi»c«/mi»«msup»«mi»m«/mi»«mn»3«/mn»«/msup»«/math» = 0.1 «math xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mi»d«/mi»«msup»«mi»m«/mi»«mn»3«/mn»«/msup»«/math»

Concentration of «math xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mi»S«/mi»«msup»«msub»«mi»O«/mi»«mn»4«/mn»«/msub»«mrow»«mn»2«/mn»«mo»-«/mo»«/mrow»«/msup»«/math» = 0.40/0.1 = 4.0 «math xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mi mathvariant=¨bold-italic¨»m«/mi»«mi mathvariant=¨bold-italic¨»o«/mi»«mi mathvariant=¨bold-italic¨»l«/mi»«mo»§nbsp;«/mo»«mi mathvariant=¨bold-italic¨»d«/mi»«msup»«mi mathvariant=¨bold-italic¨»m«/mi»«mrow»«mo»-«/mo»«mn»3«/mn»«/mrow»«/msup»«/math»

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