Ultimate Quiz On Stoichiometry Quiz

1. Mole ratios for a reaction are obtained from the

Coefficients

Chemical symbols

Molar masses

Subscripts

Mole ratios for a reaction are obtained from the coefficients. Coefficients represent the relative amounts of reactants and products involved in a chemical reaction. They are the numbers that appear in front of the chemical formulas in a balanced chemical equation. The coefficients indicate the ratio in which the substances react or are produced. By comparing the coefficients of the reactants and products, mole ratios can be determined, which are essential for stoichiometric calculations and determining the amount of each substance involved in a reaction.

Explanation

Mole ratios for a reaction are obtained from the coefficients. Coefficients represent the relative amounts of reactants and products involved in a chemical reaction. They are the numbers that appear in front of the chemical formulas in a balanced chemical equation. The coefficients indicate the ratio in which the substances react or are produced. By comparing the coefficients of the reactants and products, mole ratios can be determined, which are essential for stoichiometric calculations and determining the amount of each substance involved in a reaction.

2. Determine the mole ratio necessary to convert mole of aluminum to moles of aluminum chloride. The balanced equation is 2AL + 3Cl₂ → 2AlCl₃

2:3

2:2

3:2

The balanced equation shows that for every 2 moles of aluminum (2AL), 2 moles of aluminum chloride (2AlCl3) are formed. Therefore, the mole ratio necessary to convert moles of aluminum to moles of aluminum chloride is 2:2.

Explanation

The balanced equation shows that for every 2 moles of aluminum (2AL), 2 moles of aluminum chloride (2AlCl3) are formed. Therefore, the mole ratio necessary to convert moles of aluminum to moles of aluminum chloride is 2:2.

3. Using this balanced equation, 2C + FeCr₂O₄ → FeCr₂ + 2CO2, what mole ratio would you use to convert moles of FeCr₂O₄to moles of FeCr₂?

1:1

2:1

2:2

The given balanced equation shows that for every 2 moles of FeCr2O4, 2 moles of FeCr2 are produced. Therefore, the mole ratio to convert moles of FeCr2O4 to moles of FeCr2 is 1:1.

Explanation

The given balanced equation shows that for every 2 moles of FeCr2O4, 2 moles of FeCr2 are produced. Therefore, the mole ratio to convert moles of FeCr2O4 to moles of FeCr2 is 1:1.

4. If 5.5 mol calcium carbide (CaC₂) reacts with an excess of water, how many moles of acetylene (C₂H₂) will be produced? The balanced equation is as follows: CaC₂ + 2H₂O → Ca(OH)₂ + C₂H₂

11.0 moles

5.50 moles

352.0 moles

2.75 moles

When 5.5 moles of calcium carbide reacts with an excess of water, according to the balanced equation, it produces an equal number of moles of acetylene. Therefore, 5.5 moles of calcium carbide will produce 5.5 moles of acetylene.

Explanation

When 5.5 moles of calcium carbide reacts with an excess of water, according to the balanced equation, it produces an equal number of moles of acetylene. Therefore, 5.5 moles of calcium carbide will produce 5.5 moles of acetylene.

5. Use the following balanced equation to determine how many moles of CaCO₃ can be dissolved in 0.0250 mol of HCl. CaCO₃ + 2HCl → CaCl₂+ H₂O + CO₂

0.0250 mol

0.05 mol

0.0125 mol

2.5 mol

The balanced equation shows that 1 mole of CaCO3 reacts with 2 moles of HCl. Therefore, if 0.0250 mol of HCl is present, only half of that amount (0.0125 mol) of CaCO3 can be dissolved.

Explanation

The balanced equation shows that 1 mole of CaCO3 reacts with 2 moles of HCl. Therefore, if 0.0250 mol of HCl is present, only half of that amount (0.0125 mol) of CaCO3 can be dissolved.

6. The balanced chemical equation for the reaction that is used to fuel rockets is N₂H₂ + H2O → N₂ + 2H₂O. How many grams of N₂H₂ are needed to produce 10.0 mol nitrogen gas?

15 g

300 g

30.03 g

Option 4

In the balanced chemical equation, the stoichiometric ratio between N2H2 and N2 is 1:1. This means that for every 1 mol of N2H2, 1 mol of N2 is produced. Therefore, to produce 10.0 mol of N2, we would need an equal amount of N2H2. The molar mass of N2H2 is 30.03 g/mol, so 10.0 mol of N2H2 would weigh 10.0 mol * 30.03 g/mol = 300 g.

Explanation

In the balanced chemical equation, the stoichiometric ratio between N2H2 and N2 is 1:1. This means that for every 1 mol of N2H2, 1 mol of N2 is produced. Therefore, to produce 10.0 mol of N2, we would need an equal amount of N2H2. The molar mass of N2H2 is 30.03 g/mol, so 10.0 mol of N2H2 would weigh 10.0 mol * 30.03 g/mol = 300 g.

7. Chloroform (CHCl₃) is produced by a reaction between methane and chlorine. Use the balanced chemical equation for this reaction to determine the mass of CH₄ needed to produce 50.0 g of CHCl_3. Balanced Equation: CH₄ + 3Cl₂ → CHCl₃ + 3HCl

6.72 grams

119.37 grams

10.04 grams

11.9 grams

Based on the balanced chemical equation, 1 mole of CH4 reacts with 1 mole of CHCl3. To determine the mass of CH4 needed to produce 50.0 g of CHCl3, we need to convert grams of CHCl3 to moles using its molar mass (119.37 g/mol).

50.0 g CHCl3 * (1 mol CHCl3 / 119.37 g CHCl3) * (1 mol CH4 / 1 mol CHCl3) * (16.04 g CH4 / 1 mol CH4) = 6.72 g CH4

Therefore, the mass of CH4 needed to produce 50.0 g of CHCl3 is 6.72 grams.

Explanation

Based on the balanced chemical equation, 1 mole of CH4 reacts with 1 mole of CHCl3. To determine the mass of CH4 needed to produce 50.0 g of CHCl3, we need to convert grams of CHCl3 to moles using its molar mass (119.37 g/mol).

50.0 g CHCl3 * (1 mol CHCl3 / 119.37 g CHCl3) * (1 mol CH4 / 1 mol CHCl3) * (16.04 g CH4 / 1 mol CH4) = 6.72 g CH4

Therefore, the mass of CH4 needed to produce 50.0 g of CHCl3 is 6.72 grams.

8. Consider the following balanced equation. C₁₂H₂₂O₁₁ + 3O₂ → 2H₃C₆H₅O₇ + 3H₂O Determine the mass of citric acid (H₃C₆H₅O7) produced when 2.5 mol C₁₂H₂₂O₁₁is used.

5.0 mol

192.14 grams

481 grams

961 grams

The balanced equation shows that 1 mole of C12H22O11 produces 2 moles of H3C6H5O7. Therefore, 2.5 moles of C12H22O11 will produce 2.5 * 2 = 5 moles of H3C6H5O7.

The molar mass of H3C6H5O7 is 192.14 g/mol. Therefore, the mass of 5 moles of H3C6H5O7 is 5 * 192.14 = 960.7 grams. Rounded to the nearest whole number, the mass of citric acid produced is 961 grams.

Explanation

The balanced equation shows that 1 mole of C12H22O11 produces 2 moles of H3C6H5O7. Therefore, 2.5 moles of C12H22O11 will produce 2.5 * 2 = 5 moles of H3C6H5O7.

The molar mass of H3C6H5O7 is 192.14 g/mol. Therefore, the mass of 5 moles of H3C6H5O7 is 5 * 192.14 = 960.7 grams. Rounded to the nearest whole number, the mass of citric acid produced is 961 grams.