Đáp án:
$a,m_{AlCl_3}=53,4g.$
$b,V+_{H_2}=13,44l.$
$c,m_{ddHCl}=300g.$
$d,C\%_{AlCl_3}=17,25\%$
Giải thích các bước giải:
$a,PTPƯ:2Al+6HCl\xrightarrow{} 2AlCl_3+3H_2↑$
$n_{Al}=\dfrac{10,8}{27}=0,4mol.$
$Theo$ $pt:$ $n_{AlCl_3}=n_{Al}=0,4mol.$
$⇒m_{AlCl_3}=0,4.133,5=53,4g.$
$b,Theo$ $pt:$ $n_{H_2}=\dfrac{3}{2}n_{Al}=0,6mol.$
$⇒V_{H_2}=0,6.22,4=13,44l.$
$c,Theo$ $pt:$ $n_{HCl}=3n_{Al}=1,2mol.$
$⇒m_{HCl}=1,2.36,5=43,8g.$
$⇒m_{ddHCl}=\dfrac{43,8}{14,6\%}=300g.$
$d,m_{ddAlCl_3}=10,8+300-0,6.2=309,6g.$
$⇒C\%_{AlCl_3}=\dfrac{53,4}{309,6}.100\%=17,25\%$
chúc bạn học tốt!
