\(m_{CuSO_4}=\dfrac{8.200}{100}=16\left(g\right)\)\(\Rightarrow n_{CuSO_4}=\dfrac{16}{64}=0,25\left(mol\right)\)
\(PTHH:Fe+CuSO_4\rightarrow FeSO_4+Cu\)
Theo PTHH: \(n_{Fe}=n_{Cu}=n_{CuSO_4}=0,25\left(mol\right)\)
\(\Rightarrow m_{Fe-pu}=0,25.56=14\left(g\right)\)
\(\Rightarrow m_{Cu}=0,25.64=16\left(g\right)\)
\(\Rightarrow m_{tang}=m_{Cu}-m_{Fe}=16-14=2\left(g\right)\)
Vậy.....................
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