Đáp án:
\(a,\ m_{\text{dd HCl}}=50\ g.\\ b,\ m_{MgCl_2}=9,5\ g.\\ m_{H_2}=0,2\ g.\\ c,\ m_{\text{dd spư}}=52,2\ g.\)
Giải thích các bước giải:
\(a,\ PTHH:Mg+2HCl\to MgCl_2+H_2↑\\ n_{Mg}=\dfrac{2,4}{24}=0,1\ mol.\\ Theo\ pt:\ n_{HCl}=2n_{Mg}=0,2\ mol.\\ ⇒m_{\text{dd HCl}}=\dfrac{0,2\times 36,5}{14,6\%}=50\ g.\\ b,\ Theo\ pt:\ n_{MgCl_2}=n_{H_2}=n_{Mg}=0,1\ mol.\\ ⇒m_{MgCl_2}=0,1\times 95=9,5\ g.\\ m_{H_2}=0,1\times 2=0,2\ g.\\ c,\ m_{\text{dd spư}}=m_{Mg}+m_{\text{dd HCl}}-m_{H_2}\\ ⇒m_{\text{dd spư}}=2,4+50-0,2=52,2\ g.\)
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