Đáp án:
\(b,m=m_{Mg}=1,5\ g.\\ V=V_{H_2}=1,4\ lít.\\ c,C\%_{MgCl_2}=2,95\%\\ d,CM_{MgCl_2}=0,25M.\)
Giải thích các bước giải:
\(a,PTHH:Mg+2HCl\to MgCl_2+H_2↑\\ b,V_{HCl}=\dfrac{200}{0,8}=250\ ml=0,25\ lít.\\ ⇒n_{HCl}=0,25.0,5=0,125\ mol.\\ Theo\ pt:\ n_{Mg}=n_{H_2}=\dfrac{1}{2}n_{HCl}=0,0625\ mol.\\ ⇒m=m_{Mg}=0,0625.24=1,5\ g.\\ ⇒V=V_{H_2}=0,0625.22,4=1,4\ lít.\\ c,m_{\text{dung dịch spư}}=m_{Mg}+m_{HCl}-m_{H_2}\\ ⇒m_{\text{dung dịch spư}}=1,5+200-(0,0625.2)=201,375\ g.\\ Theo\ pt:\ n_{MgCl_2}=\dfrac{1}{2}n_{HCl}=0,0625\ mol.\\ ⇒C\%_{MgCl_2}=\dfrac{0,0625.95}{201,375}.100\%=2,95\%\\ d,CM_{MgCl_2}=\dfrac{0,0625}{0,25}=0,25M.\)
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