\(a,Zn+H_2SO_4\to ZnSO_4+H_2\\ 2Al+3H_2SO_4\to Al_2(SO_4)_3+3H_2\\ n_{H_2}=\dfrac{2,24}{22,4}=0,1(mol)\\ \Rightarrow \begin{cases} 65n_{Zn}+27n_{Al}=3,68\\ n_{Zn}+1,5n_{Al}=0,1 \end{cases}\\\Rightarrow n_{Al}=n_{Zn}=0,04(mol)\\ \Rightarrow \%m_{Al}=\dfrac{0,04.27}{3,68}.100\%=29,345\&\\ \Rightarrow \%m_{Zn}=100-29,34=70,66\%\\ b,n_{H_2SO_4}=n_{H_2}=0,1\\ \Rightarrow m_{dd_{H_2SO_4}}=\dfrac{0,1.98}{10\%}=98(g)\)
\(c,n_{ZnSO_4}=0,04(mol);n_{Al_2(SO_4)_3}=0,02(mol)\\ \Rightarrow m_{ZnSO_4}=161.0,04=6,44(g)\\ m_{Al_2(SO_4)_3}=0,02.342=6,84(g)\\ \Rightarrow C\%_{Al_2(SO_4)_3}=\dfrac{6,84}{3,68+98-0,1.2}.100\%=6,74\%\\ \Rightarrow C\%_{ZnSO_4}=\dfrac{6,44}{3,68+98-0,1.2}.100\%=6,35\%\)
2Al + 3H2SO4 -> Al2(SO4)3 + 3H2
2a 3a a 3a
Zn + H2SO4 -> ZnSO4 + H2
b b b b
nH2 = 0.1 mol
Ta có: \(\left\{{}\begin{matrix}54a+65b=3.68\\3a+b=0.1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=0.02mol\\b=0.04mol\end{matrix}\right.\)
a.\(\%mAl=\dfrac{0.02\times54\times100}{3.68}=29.35\%\)
%mZn = 100 - 29.35 = 70.65%
b.mddH2SO4\(=\dfrac{0.1\times98\times100}{10}=98g\)
Định luật bảo toàn khối lượng:
mKL + mdd ban đầu = mdd sau + mH2
<=> mdd sau = 3.68 + 98 - \(0.1\times2=101.48g\)
c.C%Al2(SO4)3 =\(\dfrac{\left(0.02\times342\right)\times100}{101.48}=6.74\%\)
C%ZnSO4 = \(\dfrac{0.04\times161\times100}{101.48}=6.35\%\)
