3,68 gam hồn hợp: \(\left\{{}\begin{matrix}Al:a\left(mol\right)\\Zn:b\left(mol\right)\end{matrix}\right.\)
\(\Rightarrow27a+65b=3,68\left(I\right)\)
\(2Al\left(a\right)+3H_2SO_4\rightarrow Al_2\left(SO_4\right)_3+3H_2\left(1,5a\right)\)
\(Zn\left(b\right)+H_2SO_4\rightarrow ZnSO_4+H_2\left(b\right)\)
\(n_{H_2}\left(đktc\right)=0,1\left(mol\right)\)
Theo PTHH: \(n_{H_2}\)thu được \(=\left(1,5a+b\right)\left(mol\right)\)
\(\Rightarrow1,5a+b=0,1\left(II\right)\)
Từ \(\left(I\right)\&\left(II\right)\Rightarrow\left\{{}\begin{matrix}a=0,04\left(mol\right)\\b=0,04\left(mol\right)\end{matrix}\right.\)
Ta có: \(n_{H_2SO_4}=1,5a+b=1,5.0,04+0,04=0,1\left(mol\right)\)
\(\Rightarrow m_{H_2SO_4}=0,1.98=9,8\left(g\right)\)\(\Rightarrow m_{ddH_2SO_4}=\dfrac{9,8.100}{10}=98\left(g\right)\)
\(m_{H_2}=0,1.2=0,2\left(g\right)\)
Khối lượng dung dịch thu được sau phản ứng là:
\(m_{ddsau}=m_{hh}+m_{ddH_2SO_4}-m_{H_2}\)
\(=3,68+98-0,2=101,48\left(g\right)\)
Chọn \(C\)
