`1,`
`n_(AgNO_3)=\frac{50.17%}{170}=0,05(mol)`
`AgNO_3+HCl->AgCl+HNO_3`
`0,05` `0,05` `0,05` `0,05`
`a,`
`m_(AgCl)=0,05.143,5=7,175(g)`
`b,`
`C%_(HCl)=\frac{0,05.36,5}{20}.100=9,125%`
`c,`
`m_(dd)=50+20-7,175=62,825(g)`
`C%_(HNO_3)=\frac{0,05.63}{62,825}.100=5,01%`
`2,`
`n_(FeCl_3)=\frac{50.16,25%}{162,5}=0,05(mol)`
`FeCl_3+3KOH->Fe(OH)_3+3KCl`
`0,05` `0,15` `0,05` `0,15`
`a,`
`m_(Fe(OH)_3)=0,05.107=5,35(g)`
`b,`
`C%_(KOH)=\frac{0,15.56}{50}.100=16,8%`
`c,`
`m_(dd)=50+50-5,35=94,65(g)`
`C5_(KCl)=\frac{0,15.74,5}{94,65}.100=11,8%`
`3,`
`n_(Cu(OH)_2)=\frac{11,76}{98}=0,12(mol)`
$Cu(OH)_2\xrightarrow{t^o}CuO+H_2O$
`0,12` `0,12`
`CuO+H_2SO_4->CuSO_4+H_2O`
`0,12` `0,12` `0,12`
`a,`
`m_(CuO)=0,12.80=9,6(g)`
`b,`
`C_(MH_2SO_4)=\frac{0,12}{0,1}=1,2M`
`c,`
`C_(MCuSO_4)=\frac{0,12}{0,1}=1,2M`