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Zigmanuir [339]
3 years ago
9

A mixture of gases is analyzed and found to have the following composition in mol %: CO2 12.0 CO 6.0 CH4 27.3 H2 9.9 N2 44.8 a)

Determine the composition of the gas mixture in weight %. b) Determine the average molecular weight of the gas mixture.
Chemistry
1 answer:
vichka [17]3 years ago
6 0

Answer:

a) CO₂: <em>21,9%; </em>CO: <em>7,0%; </em>CH₄: <em>18,2%; </em>H₂: <em>0,8%; </em>N₂: <em>52,1%</em>

b) 24,09 g/mol

Explanation:

a) It is posible to obtain the composition of the gas mixture in weight% using molecular mass of each compound, thus:

12% CO₂×\frac{44,01g}{1mol} = <em>528,1 g</em>

6% CO×\frac{28,01g}{1mol} = <em>168,1 g</em>

27,3% CH₄×\frac{16,05g}{1mol} = <em>438,2 g</em>

9,9% H₂×\frac{2,02g}{1mol} = <em>20,0 g</em>

44,8% N₂×\frac{28g}{1mol} = <em>1254,4 g</em>

The total mass of the gas mixture is:

528,1g + 168,1g + 438,2g + 20,0g + 1254,4g = <em>2408,8 g</em>

Thus composition of the gas mixture in weight% is:

CO₂: \frac{528,1g}{2408,8g}×100 = <em>21,9%</em>

CO: \frac{168,1g}{2408,8g}×100 = <em>7,0%</em>

CH₄: \frac{438,2g}{2408,8g}×100 = <em>18,2%</em>

H₂: \frac{20,0g}{2408,8g}×100 = <em>0,8%</em>

N₂: \frac{1254,4g}{2408,8g}×100 = <em>52,1%</em>

b) The average molecular weight of the gas mixture is determined with mole % composition, thus:

0,12×44,01g/mol + 0,06×28,01g/mol + 0,273×16,05g/mol + 0,099×2,02g/mol + 0,448×28g/mol = <em>24,09 g/mol</em>

I hope it helps!

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