# Ultimate Quiz On Stoichiometry Quiz

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• 1.

### Mole ratios for a reaction are obtained from the

• A.

Coefficients

• B.

Chemical symbols

• C.

Molar masses

• D.

Subscripts

A. Coefficients
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.

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• 2.

### 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

• A.

11.0 moles

• B.

5.50 moles

• C.

352.0 moles

• D.

2.75 moles

B. 5.50 moles
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.

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• 3.

### 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

• A.

0.0250 mol

• B.

0.05 mol

• C.

0.0125 mol

• D.

2.5 mol

C. 0.0125 mol
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.

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• 4.

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

• A.

5.0 mol

• B.

192.14 grams

• C.

481 grams

• D.

961 grams

D. 961 grams
Explanation
When 2.5 mol of C12H22O11 is used, according to the balanced equation, 2 mol of H3C6H5O7 is produced. To determine the mass of H3C6H5O7, we need to calculate the molar mass of H3C6H5O7. The molar mass of H3C6H5O7 is 192.14 g/mol. Therefore, when 2.5 mol of C12H22O11 is used, the mass of H3C6H5O7 produced is 2.5 mol x 192.14 g/mol = 480.35 grams. Rounded to the nearest gram, the mass of H3C6H5O7 produced is 481 grams.

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• 5.

### 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 CHCl3. Balanced Equation:  CH4 + 3Cl2 → CHCl3 + 3HCl

• A.

6.72 grams

• B.

119.37 grams

• C.

10.04 grams

• D.

11.9 grams

A. 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.

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• 6.

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

• A.

2:3

• B.

2:2

• C.

3:2

B. 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.

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• 7.

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

• A.

1:1

• B.

2:1

• C.

2:2

A. 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.

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• 8.

### 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?

• A.

15 g

• B.

300 g

• C.

30.03 g

• D.

Option 4

B. 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.

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