Question

Answer 1

Bài 5:

a) \(n_{H_2}=\dfrac{3,7185}{24,79}=0,15\left(mol\right);n_{H_2SO_4}=0,6.0,5=0,3\left(mol\right)\)

PTHH: Zn + H₂SO₄ ---> ZnSO₄ + H₂ 

          0,15<-0,15<------------------0,15

Vì 0,15 < 0,3 => H₂SO₄ dư, Zn hết

=> \(\left\{{}\begin{matrix}m_{Zn}=0,15.65=9,75\left(g\right)\\m_{Cu}=22-9,75=12,25\left(g\right)\end{matrix}\right.\)

b) \(\left\{{}\begin{matrix}C_{M\left(ZnSO_4\right)}=\dfrac{0,15}{0,6}=0,25M\\C_{M\left(H_2SO_4.dư\right)}=\dfrac{0,3-0,15}{0,6}=0,25M\end{matrix}\right.\)