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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:
fig 3.8

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}