\(n_{Al}=\frac{8,1}{27}=0,3\left(mol\right)\)
\(n_{HCl}=\frac{21,9}{36,5}=0,6\left(mol\right)\)
\(PTHH:2Al+6HCl\rightarrow2AlCl_3+3H_2\)
_______0,2_______0,6_____0,2______0,3
Lập tỉ lệ:
\(\frac{n_{Al}}{2}=0,15>\frac{n_{HCl}}{6}=0,1\)
Nên HCl pứ hết , Al dư
\(Al\left(dư\right)=Al\left(bđ\right)-Al\left(pư\right)=0,3-0,2=0,1\left(mol\right)\)
\(m_{AlCl3}=0,2.\left(27+35,5.3\right)=26,7\left(g\right)\)
\(PTHH:CuO+H_2\rightarrow Cu+H_2O\)
_______0,3_____0,3___________
\(\Rightarrow m_{CuO}=0,3.\left(64+16\right)=24\left(g\right)\)
