Question

5.6克铁粉和50克稀硫酸恰好完全反应，求反应后所得溶液的质量，溶质的质量和溶剂的质量

Answer 1

Fe+ H2SO4= FeSO4+ H25.6g x y z------------------------------------------------------------56g/mol 98g/mol 152g/mol 2g/mol算出x=9.8gy=15.2gz=0.2g反应后所得溶液的质量=5.6g+50g-0.2g=55.4g溶质的质量=15.2g溶剂的质量=55.4g-15.2g=40.2g

Answer 2

Fe+H2SO4=FeSO4+H256 152 25.6 15.2 0.2所以溶液的总质量为5.6+50-0.2=55.4g溶质的质量为15.2g溶剂的质量为55.4-15.2=40.2

Answer 3

m(溶液)=50+5.6-0.1*2=55.4gm(溶质)=56/5.6*142=14.2g(溶剂)=55.4g-14.2g=41.2g