n[SUB]AgNO[SUB]3[/SUB][/SUB]=1.0,5=0,5(mol)
n[SUB]HCl[/SUB]=0,3.2=0,6(mol)
AgNO[SUB]3[/SUB]+HCl--->AgCl+HNO[SUB]3[/SUB]
==>n[SUB]AgNO[SUB]3[/SUB][/SUB]=n[SUB]AgCl[/SUB]=0,5(mol)
==>CM[SUB]AgCl[/SUB]=\[\frac{0,5}{0,8}\]=0,625M
m[SUB]AgNO[SUB]3[/SUB][/SUB]=V.D=500.1,2=600(g)
m[SUB]HCl[/SUB]=450(g)
==>m[SUB] dd[/SUB]=1050(g)
==>C%[SUB]AgCl[/SUB]=\[\frac{0,5.143,5}{1050}.100%\]=6,8%
bạn Tính tương tự cho HNO3 và HCl dư nha:d