$m_{dd\ CuSO_4}=100.1,12=112g \\⇒m_{CuSO_4}=112.10\%=11,2g \\⇒n_{CuSO_4}=\dfrac{11,2}{160}=0,07mol \\n_{Fe}=\dfrac{1,96}{56}=0,035mol \\PTHH :$
$Fe + CuSO_4\to FeSO_4+Cu$
Theo pt : 1 mol 1 mol
Theo đbài : 0,035mol 0,07 mol
Tỉ lệ : $\dfrac{0,035}{1}<\dfrac{0,07}{1}$
⇒Sau phản ứng CuSO4 dư
Theo pt :
$n_{CuSO_4\ pư}=n_{Fe}=0,035mol \\⇒n_{CuSO_4\ dư}=0,07-0,035=0,035mol \\⇒m_{CuSO_4\ dư}=0,035.160=5,6g \\n_{FeSO_4}=n_{Cu}=n_{Fe}=0,035mol \\⇒m_{FeSO_4}=0,035.152=5,32g \\m_{Cu}=0,035.64=2,24g \\m_{dd\ spư}=1,96+112-2,24=111,72g \\⇒C\%_{CuSO_4\ dư}=\dfrac{5,6}{111,72}.100\%=5,01\% \\C\%_{FeSO_4}=\dfrac{5,32}{111,72}.100\%=4,76\%$
