Ta có: \(\left\{{}\begin{matrix}n_{H_2}=\dfrac{3,36}{22,4}=0,15\left(mol\right)\\n_{SO_2}=\dfrac{6,72}{22,4}=0,3\left(mol\right)\end{matrix}\right.\)
PTHH: \(2Al+6HCl\rightarrow2AlCl_3+3H_2\uparrow\)
0,1____0,3_____0,1_____0,15 (mol)
\(2Al+6H_2SO_{4\left(đ\right)}\xrightarrow[]{t^o}Al_2\left(SO_4\right)_3+3SO_2\uparrow+6H_2O\)
0,1______0,3__________0,05____0,15_____0,3 (mol)
\(Cu+2H_2SO_{4\left(đ\right)}\xrightarrow[]{t^o}CuSO_4+SO_2\uparrow+2H_2O\)
0,15_____0,3________0,15___0,15_____0,3 (mol)
Ta có: \(m_{Al}+m_{Cu}=0,1\cdot27+0,15\cdot64=12,3\left(g\right)\)
Bài 7:
PTHH: \(Mg+H_2SO_4\rightarrow MgSO_4+H_2\uparrow\)
Ta có: \(n_{H_2SO_4}=0,2\cdot1=0,2\left(mol\right)=n_{Mg}=n_{H_2}=n_{MgSO_4}\)
\(\Rightarrow\left\{{}\begin{matrix}m_{Ag}=15,6-0,2\cdot24=10,8\left(g\right)\\V_{H_2}=0,2\cdot22,4=2,24\left(l\right)\\C_{M_{MgSO_4}}=\dfrac{0,2}{0,2}=1\left(M\right)\end{matrix}\right.\)
