a) Zn + H2SO4 → ZnSO4 + H2↑ (1)
ZnO + H2SO4 → ZnSO4 + H2O (2)
b) \(n_{H_2}=\dfrac{2,24}{22,4}=0,1\left(mol\right)\)
Theo PT: \(n_{Zn}=n_{H_2}=0,1\left(mol\right)\)
\(\Rightarrow m_{Zn}=0,1\times65=6,5\left(g\right)\)
\(\Rightarrow m_{ZnO}=14,6-6,5=8,1\left(g\right)\)
\(\Rightarrow\%Zn=\dfrac{6,5}{14,6}\times100\%=44,52\%\)
\(\Rightarrow\%ZnO=100\%-44,52\%=55,48\%\)
c) Theo PT1: \(n_{H_2SO_4}=n_{Zn}=0,1\left(mol\right)\)
\(n_{ZnO}=\dfrac{8,1}{81}=0,1\left(mol\right)\)
theo PT2: \(n_{H_2SO_4}=n_{ZnO}=0,1\left(mol\right)\)
\(\Rightarrow\Sigma n_{H_2SO_4}=0,1+0,1=0,2\left(mol\right)\)
\(\Rightarrow C_{M_{H_2SO_4}}=\dfrac{0,2}{0,2}=1\left(M\right)\)
Theo bài ta có
nH2=2,24/22,4=0,1(mol)
Pthh ZnO+H2SO4=>ZnSO4+H2O (1)
Zn+H2SO4=>ZnSO4+H2 (2)
Theo pthh và bài ta có
nZn=nH2=0,1(mol)
=>mZn=0,1*65=6,5(g)
=>mZnO=14,6-6,5=8,1(g)
=>nZnO=8,1/81=0,1(mol)
\(\Rightarrow\left\{{}\begin{matrix}\%mZn=\dfrac{6,5\cdot100\%}{14,6}=44,52\%\\\%mZnO=100\%-44,52\%=55,48\%\end{matrix}\right.\)
Ta lại có
nH2SO4=nH2SO4(pt1)+nH2SO4(pt2)=nZn+nZnO=0,1+0,1=0,2(mol)
=>Cm( dd H2SO4)=0,2/0,2=1(M)
Good luck <3.nhớ tick cho mình nha <3