Đáp án:
$a,m=m_{Mg}=2,4g.$
$b,m_{MgCl_2}=9,5g.$
$c,V_{H_2}=2,24l.$
Giải thích các bước giải:
$a,PTPƯ:Mg+2HCl\xrightarrow{} MgCl_2+H_2↑$
$m_{HCl}=25.29,2\%=7,3g.$
$n_{HCl}=\dfrac{7,3}{36,5}=0,2mol.$
$Theo$ $pt:$ $n_{Mg}=\dfrac{1}{2}n_{HCl}=0,1mol.$
$⇒m=m_{Mg}=0,1.24=2,4g.$
$b,Theo$ $pt:$ $n_{MgCl_2}=\dfrac{1}{2}n_{HCl}=0,1mol.$
$⇒m_{MgCl_2}=0,1.95=9,5g.$
$c,Theo$ $pt:$ $n_{H_2}=\dfrac{1}{2}n_{HCl}=0,1mol.$
$⇒V_{H_2}=0,1.22,4=2,24l.$
chúc bạn học tốt!
