\(n_{H_2}=\dfrac{2,24}{2,24}=0,1\left(mol\right)\)
\(n_{H_2}=n_{H_2SO_4}=0,1\left(mol\right)\)
\(->m_{H_2SO_4}=9,8g\)
\(=>m_{ddH_2SO_4}=\dfrac{9,8.100}{10}=98\left(g\right)\)
\(=>m_{ddA}=m_{m_{ddH_2SO_4}}+m_{Al+Zn}-m_{H_2}\)
\(=98+3,68-0,2=101,48\left(g\right)\)
2Al + 3H2SO4 \(\rightarrow\)Al2(SO4)3 + 3H2 (1)
Zn + H2SO4 \(\rightarrow\)ZnSO4 + H2 (2)
nH2=\(\dfrac{2,24}{22,4}=0,1\left(mol\right)\)
Đặt nAl=a\(\Leftrightarrow\)mAl=27a
nZn=b\(\Leftrightarrow\)mZn=65b
Ta có:
\(\left\{{}\begin{matrix}27a+65b=3,68\\\dfrac{3}{2}a+b=0,1\end{matrix}\right.\)
Giải hệ pt ta có:
a=b=0,04
mdd A=0,02.342+0,04.161=13,28(g)
