1/

$m_{Ca(NO_3)_2}=0,05.164=8,2g.$

$C\%_{Ca(NO_3)_2}=\dfrac{8,2}{300}.100\%=2,73\%$

2/

$a,PTPƯ:2Al+6HCl\xrightarrow{} 2AlCl_3+3H_2↑$

$b,n_{Al}=\dfrac{10,8}{27}=0,4mol.$

$Theo$ $pt:$ $n_{H_2}=\dfrac{3}{2}n_{Al}=0,6mol.$

$⇒V_{H_2}=0,6.22,4=13,44l.$

$c,Theo$ $pt:$ $n_{AlCl_3}=n_{Al}=0,4mol.$

$⇒m_{AlCl_3}=0,4.133,5=53,4g.$

$d,Theo$ $pt:$ $n_{HCl}=3n_{Al}=1,2mol.$

$⇒V_{HCl}=\dfrac{1,2}{2}=0,6l=600ml.$

chúc bạn học tốt!

Bài 1 :

$a.m_{Ca(NO_3)_2}=0,05.164=8,2g$

$⇒C\%_{Ca(NO_3)_2}=\dfrac{8,2}{300}≈2,73\%$

Bài 2 :

$n_{Al}=10,8/27=0,4mol$

$a.2Al+6HCl\to 2AlCl_3+3H_2↑$

b.Theo pt :

$n_{H_2}=3/2.n_{Al}=3/2.0,4=0,6mol$

$⇒V_{H_2}=0,6.22,4=13,44l$

c.Theo pt :

$n_{AlCl_3}=n_{Al}=0,4mol$

$⇒m_{AlCl_3}=0,4.133,5=53,4g$

d.Theo pt :

$n_{HCl}=3.n_{Al}=3.0,4=1,2mol$

$⇒V_{HCl}=\dfrac{1,2}{2}=0,6l$