Question

标准状况下，H₂和CO的混合气体共8.96L，测得其质量为6.0g，试计算此混合气体中H₂和CO的质量和体积各为多少？

Answer 1

8.96L÷22.4L/mol=0.4mol
n_{( H2 )}×2g/mol+(0.4mol-n_{( H2 )})×28g/mol="6.0g " n_{( H2 )}=0.2mol
n_{( CO )}=0.4mol-0.2mol=0.2mol
m_{( H2 )}=0.2mol×2g/mol=0.4g
m_{( CO )}=0.2mol×28g/mol=5.6g
V_{( CO )}=V_{( H2 )}=0.2mol×22.4L/mol=4.48L

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