\(n_{H_2SO_4}=\frac{19,8}{98}=0,2\left(mol\right);n_{Zn}=\frac{6,5}{65}=0,1\left(mol\right)\)
\(PTHH:Zn+H_2SO_4\rightarrow ZnSO_4+H_2\)
(mol)_____0,1___0,1_________0,1____0,1_
Tỉ lệ: \(\frac{0,2}{1}>\frac{0,1}{1}\rightarrow H_2SO_4\) dư
\(V_{H_2}=0,1.22,4=2,24\left(l\right)\)
\(m_{H_2SO_4\cdot du}=98.\left(0,2-0,1\right)=9,8\left(g\right)\)
\(m_{ZnSO_4}=0,1.161=16,1\left(g\right)\)
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