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## 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

$N_{2}(g)$ + 3$H_{2}(g)$ → 2$NH_{3}(g)$

ⓐ 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: [$N_{2}(g)$ + 3$H_{2}(g)$ → 2$NH_{3}(g)$] moles $N_{2}$ (1.34) Information given: mol $NH_{3}$ formed Asked for:

 ANALAYSIS mass of ammonia (2.75 × 10³ g) balanced equation: [$N_{2}(g)$ + 3$H_{2}(g)$ → 2$NH_{3}$(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: [$N_{2}(g)$ + 3$H_{2}(g)$ → 2$NH_{3}$(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) + 3$H_{2}$(g) → 2$NH_{3}$(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: ## 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 $NH_{3}$