When a 16.0\text{ g} sample of pure ammonium trioxonitrate(V), NH_4NO_3, undergoes complet… — JAMB UTME

When a $16.0\text{ g}$ sample of pure ammonium trioxonitrate(V), $NH_4NO_3$ , undergoes complete thermal decomposition, it produces dinitrogen monoxide gas, $N_2O$ , and water vapor. Assuming standard temperature and pressure (STP, where molar gas volume = $22.4\text{ dm}^3\text{mol}^{-1}$ ), what is the volume of the oxide of nitrogen collected, and how does its rate of diffusion compare to that of carbon(IV) oxide, $CO_2$ , under identical conditions? (Molar masses: $N = 14\text{ g/mol}$ , $O = 16\text{ g/mol}$ , $H = 1\text{ g/mol}$ , $C = 12\text{ g/mol}$ )

$4.48\text{ dm}^3$ , and its rate of diffusion is equal to that of $CO_2$ Answer
B
$4.80\text{ dm}^3$ , and its rate of diffusion is equal to that of $CO_2$
C
$4.48\text{ dm}^3$ , and it diffuses $\sqrt{2}$ times faster than $CO_2$
D
$8.96\text{ dm}^3$ , and its rate of diffusion is slower than that of $CO_2$

Answer

4.48\text{ dm}^3

, and its rate of diffusion is equal to that of

CO_2

Thermal decomposition of

16.0\text{ g}

(

0.20\text{ mol}

) of

NH_4NO_3

produces

0.20\text{ mol}

N_2O

. At STP,

0.20\text{ mol} \times 22.4\text{ dm}^3\text{mol}^{-1} = 4.48\text{ dm}^3

. Both

N_2O

and

CO_2

have a molar mass of

44\text{ g/mol}

, so by Graham's law of diffusion, their relative rates of diffusion are identical.

Step-by-Step Solution

Write the balanced chemical equation for the thermal decomposition of ammonium trioxonitrate(V)

NH_4NO_3(s) \rightarrow N_2O(g) + 2H_2O(g)

Establishes the stoichiometric mole ratio between

NH_4NO_3

and

N_2O

, which is

1 : 1

Calculate the molar mass and number of moles of

NH_4NO_3

reacted

Molar mass of

NH_4NO_3 = 2(14) + 4(1) + 3(16) = 80\text{ g/mol}

. Number of moles =

\frac{16.0\text{ g}}{80\text{ g/mol}} = 0.20\text{ mol}

Determines the mole quantity of reactant available.

Determine the volume of

N_2O

produced at STP

Moles of

N_2O = 0.20\text{ mol}

. Volume at STP =

0.20\text{ mol} \times 22.4\text{ dm}^3\text{mol}^{-1} = 4.48\text{ dm}^3

Applies standard molar volume of gas at STP.

Compare the rates of diffusion of

N_2O

and

CO_2

using Graham's law

Molar mass of

N_2O = 2(14) + 16 = 44\text{ g/mol}

. Molar mass of

CO_2 = 12 + 2(16) = 44\text{ g/mol}

. Ratio of diffusion rates =

\sqrt{\frac{44}{44}} = 1

Gases with equal molar masses diffuse at equal rates under identical conditions.

Key Concept

Thermal decomposition of nitrogen oxides, molar volume at STP, and Graham's law of diffusion

Estimated Time:2m 0s

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