Question

有铁、铝合金11g溶于足量的稀盐酸中,在标准状况下收集到8.96L氢气,计算铁、铝合金中铁和铝各自的质量分

Answer 1

由题意：

n(Fe) + 1.5*n(Al) = 8.96/22.4 = 0.4mol

56*n(Fe) + 27*n(Al) = 11g

解得n(Fe) = 0.1mol,n(Al) = 0.2mol

所以m(Fe) = 56*0.1 = 5.6g

m(Al) = 27*0.15 = 4.05g