Ta có: \(n_{Zn}=\dfrac{13}{65}=0,2\left(mol\right)\)
a, PT: \(Zn+H_2SO_4\rightarrow ZnSO_4+H_2\)
_____0,2______0,2____________0,2 (mol)
b, \(V_{H_2}=0,2.22,4=4,48\left(l\right)\)
c, \(C_{M_{H_2SO_4}}=\dfrac{0,2}{0,4}=0,5\left(M\right)\)
d, Ta có: \(n_{CaCO_3}=\dfrac{30}{100}=0,3\left(mol\right)\)
PT: \(CaCO_3+H_2SO_4\rightarrow CaSO_4+CO_2+H_2O\)
Xét tỉ lệ: \(\dfrac{0,3}{1}>\dfrac{0,2}{1}\), ta được CaCO3 dư.
Theo PT: \(n_{CO_2}=n_{H_2SO_4}=0,2\left(mol\right)\)
\(\Rightarrow m_{CO_2}=0,2.44=8,8\left(g\right)\)