Consider the catalytic oxidation of ammonia gas represented by the balanced chemical equation below: 4NH_3(g) + 5O_2(g) \rightarrow 4NO(g) + 6H_2O(g) If 6.8\text{ g} of ammonia gas reacts completely with excess oxygen gas at STP, calculate the volume of nitrogen(II) oxide gas and the volume of steam

Consider the catalytic oxidation of ammonia gas represented by the balanced chemical equat… — JAMB UTME

Consider the catalytic oxidation of ammonia gas represented by the balanced chemical equation below:

4NH_3(g) + 5O_2(g) \rightarrow 4NO(g) + 6H_2O(g)

6.8\text{ g}

of ammonia gas reacts completely with excess oxygen gas at STP, calculate the volume of nitrogen(II) oxide gas and the volume of steam produced.

[H = 1, N = 14, O = 16, \text{Molar volume of gas at STP} = 22.4\text{ dm}^3\text{ mol}^{-1}]

Fill in the missing values in the statement below.

Answer:The volume of nitrogen(II) oxide gas (

NO

) produced at STP is 【8.96】 dm³, and the volume of steam (

H_2O

) produced at STP is 【13.44】 dm³.

Answer

The volume of nitrogen(II) oxide produced at STP is 8.96 dm³ and the volume of steam produced at STP is 13.44 dm³.

Molar mass of ammonia is 17 g/mol, meaning 6.8 g equals 0.4 mol of ammonia. According to the balanced equation, 4 moles of ammonia produce 4 moles of nitrogen(II) oxide gas and 6 moles of steam. Thus, 0.4 mol of ammonia yields 0.4 mol of nitrogen(II) oxide gas (0.4 * 22.4 = 8.96 dm³) and 0.6 mol of steam (0.6 * 22.4 = 13.44 dm³).

Step-by-Step Solution

Calculate the molar mass of ammonia (

NH_3

)

Molar mass of

NH_3 = 14 + (3 \times 1) = 17\text{ g mol}^{-1}

Required to convert the given mass of reactant into moles.

Calculate the number of moles of

NH_3

reacted

\text{Moles of } NH_3 = \frac{6.8\text{ g}}{17\text{ g mol}^{-1}} = 0.4\text{ mol}

Stoichiometric relations in chemical equations are expressed in mole ratios.

Determine the moles of

NO(g)

and

H_2O(g)

produced using stoichiometric coefficients

From the balanced equation,

4\text{ mol } NH_3 \rightarrow 4\text{ mol } NO

, so

0.4\text{ mol } NH_3 \rightarrow 0.4\text{ mol } NO

.
Also,

4\text{ mol } NH_3 \rightarrow 6\text{ mol } H_2O(g)

, so

\text{Moles of } H_2O = \frac{6}{4} \times 0.4 = 0.6\text{ mol}

The coefficients in the balanced equation define the molar ratio between reactants and products.

Convert the moles of each gaseous product to volume at STP using molar gas volume

\text{Volume of } NO = 0.4\text{ mol} \times 22.4\text{ dm}^3\text{ mol}^{-1} = 8.96\text{ dm}^3

\text{Volume of } H_2O = 0.6\text{ mol} \times 22.4\text{ dm}^3\text{ mol}^{-1} = 13.44\text{ dm}^3

At STP,

1\text{ mole}

of any ideal gas occupies

22.4\text{ dm}^3

Key Concept

Mass-Volume Stoichiometry at STP

Estimated Time:2m 0s

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