\(n_{hhA}=\frac{5,6}{22,4}=0,25\left(mol\right)\)
Zn + H2SO4 → ZnSO4 + H2↓ (1)
MgCO3 + H2SO4 → MgSO4 + CO2↑ + H2O (2)
Khí A: H2, CO2
a) Gọi x,y lần lượt là số mol của Zn và MgCO3
Ta có: \(\left\{{}\begin{matrix}65x+84y=20,05\\x+y=0,25\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=0,05\\y=0,2\end{matrix}\right.\)
Vậy \(n_{Zn}=0,05\left(mol\right)\Rightarrow m_{Zn}=0,05\times65=3,25\left(g\right)\)
\(n_{MgCO_3}=0,2\left(mol\right)\Rightarrow m_{MgCO_3}=0,2\times84=16,8\left(g\right)\)
\(\%m_{Zn}=\frac{3,25}{20,05}\times100\%=16,21\%\)
\(\%m_{MgCO_3}=100\%-16,21\%=83,79\%\)
b) \(n_{H_2SO_4}=0,6\times0,5=0,3\left(mol\right)\)
Theo PT1,2: \(\Sigma n_{H_2SO_4}pư=\Sigma n_{hhA}=0,25\left(mol\right)\)
\(\Rightarrow\Sigma n_{H_2SO_4}dư=0,3-0,25=0,05\left(mol\right)\)
\(\Rightarrow C_{M_{H_2SO_4}}dư=\frac{0,05}{0,6}=0,083\left(M\right)\)
Theo PT1: \(n_{ZnSO_4}=n_{Zn}=0,05\left(mol\right)\)
\(\Rightarrow C_{M_{ZnSO_4}}=\frac{0,05}{0,6}=0,083\left(M\right)\)
Theo PT2: \(n_{MgSO_4}=n_{MgCO_3}=0,2\left(mol\right)\)
\(\Rightarrow C_{M_{MgSO_4}}=\frac{0,2}{0,6}=0,333\left(M\right)\)
c)Theo Pt1: \(n_{H_2}=n_{Zn}=0,05\left(mol\right)\)
\(\Rightarrow m_{H_2}=0,05\times2=0,1\left(g\right)\)
Theo pT2: \(n_{CO_2}=n_{MgCO_3}=0,2\left(mol\right)\)
\(\Rightarrow m_{CO_2}=0,2\times44=8,8\left(g\right)\)
\(m_{hhA}=m_{H_2}+m_{CO_2}=0,1+8,8=8,9\left(g\right)\)
\(\Rightarrow M_{hhA}=\frac{8,9}{0,25}=35,6\left(g\right)\)
\(\Rightarrow d_{\frac{A}{H_2}}=\frac{M_A}{M_{H_2}}=\frac{35,6}{2}=17,8\)
