\(a)Mg+2HCl\rightarrow MgCl_2+H_2\\ b)n_{HCl}=0,2.1=0,2mol\\ m_{HCl}=0,2.36,5=7,3g\\ c)n_{H_2}=\dfrac{2,479}{24,79}=0,1mol\\ BTKL:m_{Mg}+m_{HCl}=m_{MgCl_2}+m_{H_2}\\ \Rightarrow m_{MgCl_2}=2,4+7,3-0,1.2=9,5g\)
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