Bài 4 :
$n_{Al}=2,3/27=0,09mol$
$a.2Al+3H_2SO_4\to Al_2(SO_4)_3+3H_2↑$
b.Theo pt :
$n_{H_2}=3/2.n_{Al}=3/2.0,09=0,135mol$
$⇒V_{H_2}=0,135.22,4=3,024l$
c.Theo pt :
$n_{H_2SO_4}=3/2.n_{Al}=3/2.0,09=0,135mol$
$⇒C_{M_{H_2SO_4}}=\dfrac{0,135}{0,1}=1,35M$
$d.m_{ddH_2SO_4}=100.1,2=120g$
Theo pt :
$n_{Al_2(SO_4)_3}=1/2.n_{Al}=1/2.0,09=0,045mol$
$⇒m_{Al_2(SO_4)_3}=0,045.342=15,39g$
$m_{dd\ spu}=2,3+120-0,135.2=122,03g$
$⇒C\%_{Al_2(SO_4)_3}=\dfrac{15,39}{122,03}.100\%=12,61\%$
Bài 5 :
$n_{H_2}=2,24/22,4=0,1mol$
$1.Fe+2HCl\to FeCl_2+H_2↑$
2.Theo pt :
$n_{Fe}=n_{H_2}=0,1mol$
$⇒m=m_{Fe}=0,1.56=5,6g$
3.Theo pt :
$n_{HCl}=2.n_{H_2}=2.0,1=0,2mol$
$⇒C_{M_{HCl}}=\dfrac{0,2}{0,2}=1M$
4.Theo pt :
$n_{FeCl_2}=n_{H_2}=0,1mol$
$⇒C_{M_{FeCl_2}}=\dfrac{0,1}{0,2}=0,5M$
