$m_{CuSO_4}=\frac{16\ \%.200}{100\ \%}=32~(g)$
$m_{CuSO_4}=\frac{32}{160}=0,2~(mol)$
$a./$
$Fe+CuSO_4→FeSO_4+Cu$
$0,2$ $0,2$ $0,2$ $0,2$ $(mol)$
$b./$
$m_{Fe}=0,2.56=11,2~(g)$
$c./$
$m_{Cu}=0,2.64=12,8~(g)$
$d./$
$m_{ct~FeSO_4}=0,2.(56+32+64)=30,4~(g)$
$m_{dd~FeSO_4}=11,2+200-12,8=198,4~(g)$
$C\ \%_{FeSO_4}=\frac{30,4. 100\ \%}{198,4}≈15,323\ \%$
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