Question

How many moles of steam are produced at 2.00 atm and 202.0°C by the complete combustion of 12.50 L of methane gas (CH4) ?

Answer 1

Answer:

1.28 mol

Explanation:

Answer 2

1.28 moles of steam are produced.

Ideal gas equation:-

$PV=nRT$....(1)

Here, P is the pressure, V is the volume, n is the number of moles, R is the gas constant, T is the temperature.

Given values:-

P=2 atm

T=202.0°C=475.15 K

V=12.50 L

R=0.08231 Latm\Kmol

Substitute all the values in the equation (1) as follows:-

$2\ atm\times12.50\ L=n\times0.0821\ \frac{Latm}{Kmol} \times475.15\ K\\\\\frac{2\ atm\times12.50\ L}{0.0821\ \frac{Latm}{Kmol} \times475.15\ K}=n\\\\0.64\ mol=n$

The reaction for the combustion of methane is as follows:-

$CH_4+2O_2\rightarrow CO_2+2H_2O$

From the above reaction:-

1 mole of $CH_4$ produces 2 moles of steam

0.64 mole of $CH_4$ produces $0.64\times2=1.28\ moles$

Hence, 1.28 moles of steam are produced.

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