Câu 20:
$n_{Mg}=\dfrac{6}{24}=0,25(mol)$
$Mg+2HCl\to MgCl_2+H_2$
$\Rightarrow n_{MgCl_2}=n_{H_2}=0,25(mol)$
$n_{HCl}=0,25.2=0,5(mol)$
$\Rightarrow m_{dd HCl}=0,5.36,5:18,25\%=100g$
$\Rightarrow m_{dd\text{spu}}=6+100-0,25.2=105,5g$
$C\%_{MgCl_2}=\dfrac{0,25.95.100}{105,5}=22,51\%$
$\to B$
Câu 21:
$n_{H_2}=\dfrac{6,72}{22,4}=0,3(mol)$
Bảo toàn nguyên tố:
$n_{Cl}=n_{HCl}=2n_{H_2}=0,6(mol)$
$\Rightarrow m_{\text{muối}}=15,8+0,6.35,5=37,1g$
$\to A$
Câu 22:
Gọi $x$ lít là thể tích dd $HCl$ $2M$
Sau khi trộn:
$V_{dd}=0,1+x (l)$
$n_{HCl}=0,1.6+2x=0,6+2x (mol)$
$\Rightarrow \dfrac{0,6+2x}{0,1+x}=3$
$\Leftrightarrow x=0,3(l)=300ml$
$\to C$
