$n_{Zn}=\frac{9,75}{65} =0,15~(mol)$
$Zn+2HCl→ZnCl_2+H_2↑$
$0,15$ $0,3$ $0,15$ $0,15$ $(mol)$
$a.)$
$V_{H_2}=0,15.22,4=3,36~(lít)$
$b.)$
$mct_{HCl}= 0,3.(35,5+1)=10,95~(g)$
$C\ \%_{HCl}=\frac{10,95.100\ \%}{100}=10,95\ \%$
$c.)$
$mct_{ZnCl_2}= 0,15.(65+71)=20,4~(g)$
$mdd_{ZnCl_2}=9,75+100-0,3=109,45~(g)$
$C\ \%_{ZnCl_2}=\frac{20,4.100\ \%}{109,45}≈18,64\ \%$
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