Chapter 3
Q. 3.9
Ammonia is used to make fertilizers for lawns and gardens by reacting nitrogen gas with hydrogen gas. The balanced equation for the reaction is
N_{2}(g) + 3H_{2}(g) → 2NH_{3}(g)
ⓐ How many moles of ammonia are formed when 1.34 mol of nitrogen react?
ⓑ How many grams of hydrogen are required to produce 2.75 × 10³ g of ammonia?
ⓒ How many molecules of ammonia are formed when 2.92 g of hydrogen react?
ⓓ How many grams of ammonia are produced when 15.0 L of air (79% by volume nitrogen) react with an excess of hydrogen? The density of nitrogen at the conditions of the reaction is 1.25 g/L.
STRATEGY
For all parts of this example, follow the schematic pathway shown in Figure 3.8.
ⓐ
| ANALAYSIS | |
| balanced equation: [N_{2}(g) + 3H_{2}(g) → 2NH_{3}(g)] moles N_{2} (1.34) |
Information given: |
| mol NH_{3} formed | Asked for: |
ⓑ
| ANALAYSIS | |
| mass of ammonia (2.75 × 10³ g) balanced equation: [N_{2}(g) + 3H_{2}(g) → 2NH_{3}(g)] |
Information given: |
| molar masses of NH_{3} and H_{2} | Information implied: |
| mass of H_{2} needed | Asked for: |
ⓒ
| ANALAYSIS | |
| mass of H_{2} (2.92 g) balanced equation: [N_{2}(g) + 3H_{2}(g) → 2NH_{3}(g)] |
Information given: |
| molar mass of H_{2} Avogadro’s number (N_{A}) |
Information implied: |
| molecules of NH_{3} produced | Asked for: |
ⓓ
| ANALAYSIS | |
| balanced equation: [N_{2}(g) + 3H_{2}(g) → 2NH_{3}(g)] V_{air} (15.0 L); % by volume of N_{2} in air (79%); density, d, of N_{2} (1.25 g/L) |
Information given: |
| molar masses of N_{2} and NH_{3} | Information implied: |
| mass of NH_{3} produced | Asked for: |
Step-by-Step
Verified Solution
ⓐ
| mol H_{2} → mol NH_{3}
1.34 mol H_{2} × \frac{2 mol NH_{3}}{1 mol N_{2}} = 2.68 mol NH_{3} |
mol NH_{3} |
ⓑ
| mass NH_{3} → mol NH_{3} → mol H_{2} → mass H_{2}
2.75 × 10³ g NH_{3} × \frac{1 mol NH_{3}}{17.03 g NH_{3}} × \frac{3 mol H_{2}}{2 mol NH_{3}} × \frac{2.016 g H_{2}}{1 mol H_{2}} = 488 g H_{2} |
mass H_{2} |
ⓒ
| mass mass H_{2} \rightarrow mol H_{2} \rightarrow mol NH_{3} \underrightarrow{N_{A}} molecules NH_{3}
2.92 g H_{2} × \frac{1 mol H_{2}}{2.016 g H_{2}} × \frac{2 mol NH_{3}}{3 mol H_{2}} × \frac{6.022 × 10^{23}molecules}{1 mol NH_{3}} = 5.81 × 10^{23} |
molecules NH_{3} |
ⓓ
| V_{air} \underrightarrow{\%N_{2} } V_{N_{2}}\underrightarrow{density} mass N_{2} \rightarrow mol N_{2} \rightarrow mol NH_{3} \rightarrow mass NH_{3}
15.0 L air × \frac{79 L N_{2}}{100 L air} × \frac{1.25 g N_{2}}{1 L N_{2}} × \frac{1 mol N_{2}}{28.02 g N_{2}} × \frac{2 mol NH_{3}}{1 mol N_{2}} × \frac{17.03 g NH_{3}}{1 mol NH_{3}} = 18 g NH_{3} |
mass NH_{3} |