Question

取铜银合金1G放入硝酸溶液中溶解,再加盐酸,则得氯化银沉淀0.27G,求此合金中铜.银的质量分数

Answer 1

n(Ag)=n(AgCl)=0.27/143.5 mol=0.002mol

m(Ag)=108*0.002 g=0.216g

银的质量分数:0.216/1*100%=21.6%

铜的质量分数:1-21.6%=78.4%