Ch (2) : Calculations With Chemical Formulas And Equations

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Ch (2) : Calculations With Chemical Formulas And Equations - Quiz

QUIZ FOR (( calculations with chemical formulas and equations ))
Faculty of Engineering, Zagazig University
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Questions and Answers
  • 1. 

    Percentage of Magnesium [Mg] in Magnesium sulfate [MgSO4] Given Atomic Weights:  Mg = 24  .   S = 32   .  O = 16  

    • A.

      40

    • B.

      20

    • C.

      12

    • D.

      60

    Correct Answer
    A. 40
    Explanation
    The percentage of Magnesium (Mg) in Magnesium sulfate (MgSO4) can be calculated by dividing the atomic weight of Magnesium by the molar mass of Magnesium sulfate and multiplying by 100. The molar mass of Magnesium sulfate (MgSO4) can be calculated by adding the atomic weights of Magnesium (Mg), Sulfur (S), and Oxygen (O). In this case, the atomic weight of Magnesium is 24, and the molar mass of Magnesium sulfate is 24 + (32 + (4 * 16)) = 120. Therefore, the percentage of Magnesium in Magnesium sulfate is (24 / 120) * 100 = 20%.

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  • 2. 

    Balance the following equation: Mg3N2 + H2O  Mg(OH)2 + NH3

    • A.

      Mg3N2 + 6H2O --> 3Mg(OH)2 + 2NH3

    • B.

      Mg3N2 + 6H2O --> 3Mg(OH)2 + NH3

    • C.

      Mg3N2 + 2H2O --> Mg(OH)2 + NH3

    • D.

      Mg3N2 + 3H2O --> 3Mg(OH)2 + 2NH3

    Correct Answer
    A. Mg3N2 + 6H2O --> 3Mg(OH)2 + 2NH3
    Explanation
    The balanced equation shows that for every 1 molecule of Mg3N2, 6 molecules of H2O are required to produce 3 molecules of Mg(OH)2 and 2 molecules of NH3. This is achieved by adjusting the coefficients in front of each compound to ensure that the number of atoms of each element is the same on both sides of the equation.

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  • 3. 

    Calculate the mass % of carbon in carbon monoxide.

    Correct Answer
    42.84
    42.8
    43
    Explanation
    The correct answer is 42.84, 42.8, 43. The mass % of carbon in carbon monoxide can be calculated by dividing the mass of carbon by the total mass of the compound and multiplying by 100. In carbon monoxide, the molar mass of carbon is 12.01 g/mol and the molar mass of carbon monoxide is 28.01 g/mol. Therefore, the mass % of carbon in carbon monoxide is (12.01 g/mol / 28.01 g/mol) * 100 = 42.84%. The other options, 42.8% and 43%, are also within a reasonable range and can be considered correct.

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  • 4. 

    A reaction that produces crude iron from iron ore is shown below: Fe2O3 (s) + 3CO (g 2Fe (s) + 3CO2 (g) How many moles of iron could be produced from the reaction of 10 mol Fe2O3 and 25 mol of CO?   

    • A.

      10 mol

    • B.

      25 mol

    • C.

      17 mol

    • D.

      20 mol

    • E.

      35 mol

    Correct Answer
    C. 17 mol
    Explanation
    From the balanced equation, we can see that the stoichiometric ratio between Fe2O3 and Fe is 2:2, meaning that for every 2 moles of Fe2O3, we will produce 2 moles of Fe. Therefore, if we have 10 moles of Fe2O3, we will produce 10 moles of Fe. The stoichiometric ratio between CO and Fe is 3:2, meaning that for every 3 moles of CO, we will produce 2 moles of Fe. Therefore, if we have 25 moles of CO, we will produce 25*(2/3) = 16.67 moles of Fe. Since we cannot have a fraction of a mole, the maximum number of moles of Fe that could be produced from the given reactants is 17 mol.

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  • 5. 

    You are given the following percentages: 40.05% S and 59.95% O. Find the empirical formula for these elements

    • A.

      SO

    • B.

      SO2

    • C.

      SO4

    • D.

      SO3

    Correct Answer
    D. SO3
    Explanation
    The empirical formula is determined by the simplest whole number ratio of atoms in a compound. In this case, the percentages given suggest that there is a 40.05% chance of finding S and a 59.95% chance of finding O in the compound. To find the simplest ratio, we divide both percentages by the smallest percentage value, which is 40.05%. When we do this, we get approximately 1 S and 1.5 O. To simplify further, we multiply both values by 2 to get 2 S and 3 O. Therefore, the empirical formula for these elements is SO3.

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  • 6. 

    Given the following: 42.07% Na, 18.89% P, and 39.04% O, determine the empirical formula

    • A.

      NaPO2

    • B.

      Na2PO3

    • C.

      Na3PO4

    • D.

      NaPO4

    Correct Answer
    C. Na3PO4
    Explanation
    The empirical formula is determined by finding the ratio of the different elements in the compound. In this case, we have 42.07% Na, 18.89% P, and 39.04% O. To find the ratio, we convert these percentages to moles. Then, we divide the moles of each element by the smallest number of moles to get the simplest ratio. In this case, we find that the ratio is 1:1:4 for Na, P, and O respectively. Therefore, the empirical formula is Na3PO4.

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  • 7. 

    Styrene has the empirical formula CH, and there is 92.25% carbon and 7.75% hydrogen. If you have a molar mass of 104g/mol, what is the multiplier or factor to get the molecular formula?

    • A.

      6

    • B.

      2

    • C.

      4

    • D.

      8

    Correct Answer
    D. 8
    Explanation
    The empirical formula of styrene is CH, which means it contains one carbon atom and one hydrogen atom. The molar mass of styrene is given as 104g/mol. To find the molecular formula, we need to determine the actual number of atoms in the molecule.

    We are given that styrene is 92.25% carbon and 7.75% hydrogen. We can assume that the percentages represent the mass percentages of each element in the compound.

    To find the number of carbon atoms, we divide the mass percentage of carbon by the molar mass of carbon (12g/mol) and multiply by the molar mass of the compound (104g/mol).

    (92.25% carbon / 12g/mol) * 104g/mol = 773 atoms of carbon

    To find the number of hydrogen atoms, we divide the mass percentage of hydrogen by the molar mass of hydrogen (1g/mol) and multiply by the molar mass of the compound (104g/mol).

    (7.75% hydrogen / 1g/mol) * 104g/mol = 806 atoms of hydrogen

    The ratio of carbon to hydrogen atoms is approximately 773:806. Simplifying this ratio gives us 1:1.

    Therefore, the multiplier or factor to get the molecular formula is 1. However, the correct answer provided is 8, which is incorrect.

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  • 8. 

    Al(OH)3 + 3HCl --> 3H2O + AlCl3 If you have 14.0 grams of aluminum hydroxide, how much aluminum chloride is produced?

    • A.

      14.0 g

    • B.

      17.9 g

    • C.

      23.9 g

    Correct Answer
    C. 23.9 g
    Explanation
    When aluminum hydroxide (Al(OH)3) reacts with hydrochloric acid (HCl), it forms water (H2O) and aluminum chloride (AlCl3). The balanced equation shows that 1 mole of Al(OH)3 reacts with 3 moles of HCl to produce 3 moles of AlCl3. To determine the amount of AlCl3 produced, we need to convert the given mass of Al(OH)3 to moles, using its molar mass. Then, using the mole ratio from the balanced equation, we can calculate the moles of AlCl3 produced. Finally, we convert the moles of AlCl3 to grams, using its molar mass, to find that 23.9 grams of aluminum chloride is produced.

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  • 9. 

    Al(OH)3 + 3HCl --> 3H2O + AlCl3 If you actually recovered 22.0 grams of aluminum chloride, what is the percent yield of the reaction?

    • A.

      100%

    • B.

      92%

    • C.

      69%

    Correct Answer
    B. 92%
    Explanation
    The percent yield of a reaction is calculated by dividing the actual yield (in this case, 22.0 grams of aluminum chloride) by the theoretical yield (the maximum amount of aluminum chloride that could be produced based on the stoichiometry of the balanced equation) and multiplying by 100. Since the equation is balanced with a 1:1 ratio between aluminum chloride and aluminum hydroxide, the theoretical yield of aluminum chloride would also be 22.0 grams. Therefore, the percent yield is (22.0/22.0) x 100 = 100%, which matches the first option.

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  • 10. 

    What is the percentage of oxygen in carbon dioxide? (CO2)

    Correct Answer
    72.7%
    72.7
    72
    Explanation
    The percentage of oxygen in carbon dioxide (CO2) is 72.7%. This means that out of the total composition of CO2, 72.7% is made up of oxygen.

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  • 11. 

    Empirical Formula: 39.8% K, 27.8% Mn, 32.5% O

    • A.

      N3,Mn,5O

    • B.

      K2,Mn,O4

    • C.

      Mn3,K,O7

    • D.

      O10,Mn,5K

    Correct Answer
    B. K2,Mn,O4
    Explanation
    The empirical formula represents the simplest ratio of atoms in a compound. In this case, the compound contains 39.8% K, 27.8% Mn, and 32.5% O. To determine the empirical formula, we need to find the ratio of these elements. By dividing the percentages by their respective atomic masses and then dividing by the smallest value obtained, we find that the ratio is approximately 2:1:4. Therefore, the empirical formula is K2MnO4.

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  • 12. 

    The simplest formula of a substance shows

    • A.

      the actual number of atoms of each element in one molecule of a substance.

    • B.

      the elements that make up one molecule of the substance and the simplest whole number ratio between the atoms.

    • C.

      the number of molecules in a sample of the substance.

    • D.

      the molecular mass of the substance.

    Correct Answer
    B. the elements that make up one molecule of the substance and the simplest whole number ratio between the atoms.
    Explanation
    The simplest formula of a substance shows the elements that make up one molecule of the substance and the simplest whole number ratio between the atoms. This means that the formula provides information about the types of atoms present in the molecule and how they are arranged in the simplest ratio. It does not provide information about the actual number of atoms or molecules in a sample of the substance, or the molecular mass of the substance.

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  • 13. 

     A compound is found to have a molecular mass of 90 atomic mass units and simplest formula of C2H5O. The molecular formula of the substance is: (Atomic mass of C = 12 amu, H = 1 amu, O = 16 amu)

    • A.

      C3H6O3

    • B.

      C4H26O

    • C.

      C4H10O2

    • D.

      C5H14O

    Correct Answer
    C. C4H10O2
    Explanation
    The molecular mass of the compound is 90 atomic mass units. The simplest formula of the compound is C2H5O. To find the molecular formula, we need to determine the ratio between the molecular mass and the simplest formula mass. The simplest formula mass can be calculated by adding the atomic masses of each element in the formula: (2 * 12 amu) + (5 * 1 amu) + (1 * 16 amu) = 46 amu. Dividing the molecular mass (90 amu) by the simplest formula mass (46 amu) gives us a ratio of approximately 1.96. Multiplying the subscripts in the simplest formula by this ratio, we get C4H10O2 as the molecular formula.

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  • 14. 

    How many grams of oxygen are in one mole of calcium carbonate, CaCO3? (Atomic mass of O = 16 amu.)

    • A.

      16 grams

    • B.

      32 grams

    • C.

      48 grams

    • D.

      64 grams

    Correct Answer
    C. 48 grams
    Explanation
    Calcium carbonate (CaCO3) consists of one calcium atom (Ca), one carbon atom (C), and three oxygen atoms (O). The atomic mass of oxygen is 16 amu. To find the total mass of oxygen in one mole of calcium carbonate, we need to multiply the atomic mass of oxygen by the number of oxygen atoms in the compound (3). Therefore, the total mass of oxygen in one mole of calcium carbonate is 16 amu * 3 = 48 grams.

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  • 15. 

    The ionic compound containing Fe3+ and SO42- would have the formula

    • A.

      FeSO4

    • B.

      Fe2SO4

    • C.

      Fe2(SO4)3

    • D.

      Fe3(SO4)2

    Correct Answer
    C. Fe2(SO4)3
    Explanation
    The correct answer is Fe2(SO4)3. This is because Fe3+ has a charge of +3 and SO42- has a charge of -2. In order for the compound to be neutral, the charges must balance out. Therefore, two Fe3+ ions are needed to balance out the charge of three SO42- ions, resulting in the formula Fe2(SO4)3.

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