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

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

【解析】略