Question

mn giải giúp mik vs nha

Answer 1

\(n_{CuSO_4}=0,2x\left(mol\right)\)

\(Fe+CuSO_4\rightarrow FeSO_4+Cu\)

0,2x    0,2x                         0,2x

\(m_{tăng}=m_{Cu}-m_{Fe}=64\cdot0,2x-56\cdot0,2x=1,6\)

\(\Rightarrow x=1M\)

Chọn C.

Answer 2

\(n_{Fe}=n_{FeSO_4}=n_{Cu}=n_{CuSO_4}=0,2.x\left(mol\right)\\ Fe+CuSO_4\rightarrow FeSO_4+Cu\\ m_{t\text{ăn}g}=m_{Cu.b\text{á}m.v\text{ào}}-m_{Fe.tan.ra}\\ \Leftrightarrow1,6=64.0,2x-56.0,2x\\ \Leftrightarrow x=1\\ \Rightarrow C\)