$a)PTHH:2Al+6HCl\to 2AlCl_3+3H_2$
$b)n_{Al}=\dfrac{5,4}{27}=0,2(mol)$
Theo PT: $n_{H_2}=1,5.n_{Al}=0,3(mol)$
$\Rightarrow V_{H_2}=0,3.22,4=6,72(lít)$
$c)$Theo PT: $n_{HCl}=3n_{Al}=0,6(mol)$
$\Rightarrow m_{HCl}=0,6.36,5=21,9(g)$
$d)PTHH:2H_2+O_2\xrightarrow{t^o}2H_2O$
Theo PT: $n_{H_2O}=n_{H_2}=0,3(mol)$
$\Rightarrow m_{H_2O}=0,3.18=5,4(g)$
2Al+6HCl->2AlCl3+3H2
0,2--0,6----------------0,3
2H2+O2-to>2H2O
0,3-------------0,3
=>n Al=\(\dfrac{5,4}{27}\)=0,2 mol
=>VH2=0,3.22,4=6,72l
=>m HCl=0,6.26,5=21,9g
=>m H2O=0,3.18=5,4g
\(a,PTHH:2Al+6HCl\rightarrow2AlCl_3+3H_2\)
\(b,n_{Al}=\dfrac{m}{M}=\dfrac{5,4}{27}=0,2\left(mol\right)\\ Theo.PTHH:n_{H_2}=\dfrac{3}{2}.n_{Al}=\dfrac{3}{2}.0,2=0,3\left(mol\right)\\ V_{H_2\left(đktc\right)}=n.22,4=0,3.22,4=6,72\left(l\right)\)
\(c,Theo.PTHH:n_{HCl}=3.n_{Al}=3.0,2=0,6\left(mol\right)\\ m_{HCl}=n.M=0,6.36,5=21,9\left(g\right)\)
\(d,PTHH:2H_2+O_2\underrightarrow{t^o}2H_2O\\ Theo.PTHH:n_{H_2O}=n_{H_2}=0,3\left(mol\right)\\ m_{H_2O}=n.M=0,3.18=5,4\left(g\right)\)
