1.
How many moles of H_{2}O will be produced from the complete combustion of 2.4 grams of CH_{4}?
CH_{4} + 2O_{2} Â® CO_{2} + 2H_{2}O
Correct Answer
B. 0.30
Explanation
The balanced equation shows that for every 1 mole of CH4, 2 moles of H2O are produced. To find the number of moles of H2O produced, we need to convert the mass of CH4 to moles. The molar mass of CH4 is 16.04 g/mol. Therefore, 2.4 g of CH4 is equal to 2.4 g / 16.04 g/mol = 0.15 mol. Since the ratio of CH4 to H2O is 1:2, the number of moles of H2O produced is 0.15 mol x 2 = 0.30 moles.
2.
If 25.0 g of each reactant were used in performing the following reaction, which would be the limiting reactant?
3PbO_{2} + Cr_{2}(SO_{4})_{3} + K_{2}SO_{4} + H_{2}O Â® 3PbSO_{4} + K_{2}Cr_{2}O_{7} + H_{2}SO_{4}
Correct Answer
A. PbO2
Explanation
The given reaction equation shows that 3 moles of PbO2 are required to produce 3 moles of PbSO4. If 25.0 g of PbO2 is used, we can calculate the number of moles by dividing the mass by the molar mass of PbO2. Similarly, we can calculate the number of moles for each reactant. Comparing the number of moles of PbO2 to the other reactants, we find that PbO2 has the lowest number of moles. Therefore, PbO2 is the limiting reactant in this reaction.
3.
If you have 4.5 moles of lead(IV) nitrate howmay moles of silver nitrate can be produced according to the following reaction?
Pb(NO3)4 + 4AgCl ---> 4AgNO3 + PbCl4
Correct Answer
D. 18n
Explanation
In the given reaction, 1 mole of lead(IV) nitrate (Pb(NO3)4) reacts with 4 moles of silver chloride (AgCl) to produce 4 moles of silver nitrate (AgNO3) and 1 mole of lead(IV) chloride (PbCl4). Therefore, the mole ratio between lead(IV) nitrate and silver nitrate is 1:4. Since we have 4.5 moles of lead(IV) nitrate, we can multiply this by the mole ratio to find the number of moles of silver nitrate produced. 4.5 moles of lead(IV) nitrate multiplied by 4 gives us 18 moles of silver nitrate. Hence, the correct answer is 18n.
4.
Starting with 31g of Al in the following reaction will produce hao many grams of Na?
Na3PO4 + Al ---> AlPO4 + 3Na
Correct Answer
C. 79g
Explanation
In the given reaction, 1 mole of Al reacts with 1 mole of Na3PO4 to produce 1 mole of AlPO4 and 3 moles of Na. The molar mass of Al is 26.98 g/mol and the molar mass of Na is 22.99 g/mol.
To find the grams of Na produced, we need to calculate the moles of Al first.
Using the formula: moles = mass/molar mass, we can calculate the moles of Al:
moles of Al = 31g/26.98 g/mol = 1.15 mol
Since the mole ratio between Al and Na is 1:3, the moles of Na produced will be 3 times the moles of Al:
moles of Na = 1.15 mol * 3 = 3.45 mol
Finally, we can calculate the grams of Na produced using the formula:
grams of Na = moles of Na * molar mass of Na = 3.45 mol * 22.99 g/mol = 79g.
Therefore, the correct answer is 79g.
5.
If you counted 6.0 x 10^24 molecules of hydrogen gas how many molecules of oxygen gas will be need to completely react the amount according to the following equation?
2H2 + O2 ---> 2H2O
Correct Answer
C. 3.0 x 10^24
Explanation
To completely react the given amount of hydrogen gas, according to the balanced equation, we need an equal number of molecules of oxygen gas. Therefore, the correct answer is 3.0 x 10^24.
6.
40.0g of Sugar(C6H12O6) will product what volume of carbon dioxide according to the following equation?
C6H12O6 + 6O2 ---> 6H2O + 6CO2
Correct Answer
D. 29.9L
Explanation
When 40.0g of sugar (C6H12O6) reacts with oxygen (O2) according to the given balanced equation, it produces 6 moles of carbon dioxide (CO2). Using the molar mass of sugar (180.16 g/mol) and the molar volume of a gas at standard temperature and pressure (22.4 L/mol), we can calculate the volume of carbon dioxide produced. The calculation is as follows:
40.0g C6H12O6 * (1 mol C6H12O6 / 180.16 g C6H12O6) * (6 mol CO2 / 1 mol C6H12O6) * (22.4 L / 1 mol CO2) = 29.9L
Therefore, the correct answer is 29.9L.
7.
Using the following equation how many liters of hydrogen gas will be needed to react with 5.0L of oxygen gas?
2H2 + O2 ---> 2H2O
Correct Answer
C. 10.L
Explanation
The balanced equation shows that 2 moles of hydrogen gas react with 1 mole of oxygen gas to produce 2 moles of water. Therefore, the ratio of hydrogen gas to oxygen gas is 2:1. Since the question states that 5.0L of oxygen gas is given, we can use this ratio to determine the amount of hydrogen gas needed. Since the ratio is 2:1, we would need twice the amount of oxygen gas, which is 10.0L. Therefore, the correct answer is 10.0L.
8.
Starting with 20.5L of hydrogen gas how many molecules of amonia can be made according to this equation?
N2 + 3H2 ---> 2NH3
Correct Answer
B. 3.67 x 10^23
Explanation
According to the balanced equation, 1 molecule of N2 reacts with 3 molecules of H2 to produce 2 molecules of NH3. Therefore, the number of molecules of NH3 that can be produced is directly proportional to the number of molecules of H2.
Starting with 20.5L of H2 gas, we can use the ideal gas law to convert the volume of gas to the number of molecules.
Using Avogadro's law, we know that 1 mole of any gas contains 6.022 x 10^23 molecules. The molar volume of any gas at standard temperature and pressure (STP) is 22.4L.
Therefore, the number of moles of H2 gas is calculated as (20.5L / 22.4L) = 0.915 moles.
Since the balanced equation shows that 3 moles of H2 are required to produce 2 moles of NH3, we can calculate the number of moles of NH3 produced as (0.915 moles H2) x (2 moles NH3 / 3 moles H2) = 0.61 moles NH3.
Finally, we can convert the moles of NH3 to molecules using Avogadro's number: (0.61 moles NH3) x (6.022 x 10^23 molecules / 1 mole) = 3.67 x 10^23 molecules of NH3.
Therefore, the correct answer is 3.67 x 10^23.
9.
If starting with 1.2 x 10^24 atoms of potassium how many grams of potassium oxide can you produce using the following equation?
4K + O2 ---> 2K2O
Correct Answer
D. 94g
Explanation
The molar mass of potassium is 39.1 g/mol and the molar mass of potassium oxide is 94 g/mol. According to the balanced equation, 4 moles of potassium react with 1 mole of oxygen to produce 2 moles of potassium oxide. Therefore, the ratio of moles of potassium to moles of potassium oxide is 4:2 or 2:1. This means that for every 2 moles of potassium oxide produced, we need 4 moles of potassium. Given that we start with 1.2 x 10^24 atoms of potassium, we can convert this to moles using Avogadro's number. Finally, we can use the molar mass of potassium oxide to convert moles to grams. The result is 94g.
10.
3.50 x 10^25 compounds of calcium carbonate will produce what volume of carrbon dioxide according to the following equation?
CaCO3 ---> CaO + CO2
Correct Answer
C. 1300L
Explanation
The equation provided shows that 1 mole of calcium carbonate produces 1 mole of carbon dioxide. Since the question states that there are 3.50 x 10^25 compounds of calcium carbonate, we can assume that there are also 3.50 x 10^25 moles of calcium carbonate. Therefore, according to the stoichiometry of the equation, there will be 3.50 x 10^25 moles of carbon dioxide produced. To convert moles to liters, we need to use the ideal gas law. Assuming standard temperature and pressure, 1 mole of any gas occupies 22.4 liters. Therefore, 3.50 x 10^25 moles of carbon dioxide will occupy 3.50 x 10^25 * 22.4 liters, which is approximately 1300 liters.
11.
If 25g of NaCl is produced by the following reaction how many atoms of sodium would be required?2Na + Cl2 ---> 2NaCl
Correct Answer
B. 2.6 x 10^23
Explanation
In the given reaction, 2 moles of sodium (2Na) react with 1 mole of chlorine (Cl2) to produce 2 moles of sodium chloride (2NaCl). Since 1 mole of any substance contains Avogadro's number (6.022 x 10^23) of atoms or molecules, we can say that 2 moles of sodium will contain 2 times Avogadro's number of sodium atoms. Therefore, the correct answer is 2.6 x 10^23 atoms of sodium, as it represents 2 times Avogadro's number rounded to one significant figure.
12.
When reacting 448L of hydrogen gas with oxygen gas how many grams of water can be produced?2H2 + O2 ---> 2H2O
Correct Answer
A. 360 g
Explanation
In the balanced chemical equation, it is stated that 2 moles of hydrogen gas react with 1 mole of oxygen gas to produce 2 moles of water. The molar mass of hydrogen gas is 2 g/mol, and the molar mass of water is 18 g/mol. By using the given reaction stoichiometry and the molar masses, we can calculate the amount of water produced.
448 L of hydrogen gas can be converted to moles using the ideal gas law. Then, using the mole ratio from the balanced equation, we can determine the moles of water produced. Finally, multiplying the moles of water by the molar mass of water gives the mass of water produced, which is 360 g.