氨气与硫酸反应
2NH3 + H2SO4 = (NH4)2SO4
2 1
n 1mol×0.1l=0.1mol
2/n = 1/1
n = 0.2mol
由氮原子守恒,知氯化铵的物质的量为0.2mol
则氯化铵的质量分数为:0.2mol×53.5g/mol÷11g×100%=97.3%
(2)氨气溶于水,CNH3 = 0.2mol/0.1l = 2mol/l
氨气与硫酸反应
2NH3 + H2SO4 = (NH4)2SO4
2 1
n 1mol×0.1l=0.1mol
2/n = 1/1
n = 0.2mol
由氮原子守恒,知氯化铵的物质的量为0.2mol
则氯化铵的质量分数为:0.2mol×53.5g/mol÷11g×100%=97.3%
(2)氨气溶于水,CNH3 = 0.2mol/0.1l = 2mol/l