Question

Cho 2,7g Al vào 100g dung dịch H2SO4

a) Tính VH2 (đktc)

b) Tính C% H2SO4

c) Tính C% dung dịch sau pư

Answer 1

n Al=2,7/27=0,1(mol)

2Al + 3H2SO4 ---> Al2(SO4)3 + 3H2

0,1...........0,15..................0,05..........0,15

V H2= 0,15.22,4=3,36(l)

m H2SO4= 0,15.98=14,7(g)

C% H2SO4= 14,7.100%/100=14,7%

m dd sau phản ứng = 2,7 + 100 - 0,15.2=102,4(g)

m Al2(SO4)3 = 0,05.342 =17,1(g)

C% = 17,1.100%/102,4=16,7%

Answer 2

$n_{Al}=2,7/27=0,1mol$

$PTHH :$

$2Al+3H_2SO_4\to Al_2(SO_4)_3+3H_2$

a.Theo pt :

n_{H_2}=3/2.n_{Al}=3/2.0,2=0,15mol

$=>V_{H_2}=0,15.22,4=3,36l

b.Theo pt :

$n_{H_2SO_4}=3/2.n_{Al}=3/2.0,1=0,15mol$

$=>m_{H_2SO_4}=0,15.98=14,7g$

$=>C\%_{H_2SO_4}=\dfrac{14,7}{100}.100\%=14,7\%$

c.Theo pt :

$n_{H_2}=3/2.n_{Al}=3/2.0,1=0,15mol$

$n_{Al_2(SO_4)_3}=1/2.n_{Al}=1/2.0,1=0,05mol$

$=>m_{Al_2(SO_4)_3}=0,05.342=17,1g$

$m_{dd\ spu}=2,7+100-0,15.2=102,4g$

$\Rightarrow C\%_{Al_2(SO_4)_3}=\dfrac{17,1}{102,4}.100\%=16,7\%$