Đáp án:

$$\downarrow$$

Giải thích các bước giải:

 n_(Al)=(5,4)/27=0,2(mol)

a, @ PTHH:   2Al+3H_2SO_4->Al_2(SO_4)_3+3H_2

->n_(H_2)=3/2 n_(Al)=0,3(mol)

->V_(H_2( đktc))=0,3xx22,4=6,72(l)

   V_(H_2(đkc))=0,3xx24,79=7,437(l)

b, n_(H_2SO_4)=n_(H_2)=0,3(mol)

->m_(H_2SO_4)=0,3xx98=29,4(g)

->C%_(H_2SO_4)=(29,4)/150 xx100% = 19,6%

c, n_(Al_2(SO_4)_3)=1/2 n_(Al)=0,1(mol)

m_(dd\ sau)=5,4+150-0,3xx2=154,8(g)

->C%_(Al_2(SO_4)_3)=(0,1xx342)/(154,8) xx100% ~~22,09%

#Kudo

n_{Al}={5,4}/{27}=0,2(mol)

2Al+3H_2SO_4->Al_2(SO_4)_3+3H_2

a)

Theo PT: n_{H_2}=3/2n_{Al}=0,3(mol)

->V_{H_2}=0,3.22,4=6,72(L)

b)

Theo PT: n_{H_2SO_4}=n_{H_2}=0,3(mol)

->C\%_{H_2SO_4}={0,3.98}/{150}.100\%=19,6\%

c)

Theo PT: n_{Al_2(SO_4)_3}=1/2n_{Al}=0,1(mol)

m_{dd\ spu}=5,4+150-0,3.2=154,8(g)

->C\%_{Al_2(SO_4)_3}={0,1.342}/{154,8}.100\%\approx 22,09\%