Question

Answer 1

Câu 5: 

nH2= 5,6/22,4=0,25(mol)

PTHH: Fe + H2SO4 -> FeSO4 + H2

nH2SO4=nH2=0,25(mol)

=> VddH2SO4= 0,25/1=0,25(l)=250(ml)

=>V=250(ml)

Answer 2

Bài 6:

PTHH: Zn +2 HCl -> ZnCl2 + H2

nH2=6,72/22,4=0,3(mol)

Ta có: nZn=nH2=0,3(mol)

=>mZn=0,3.65=19,5(g)

=>m=19,5(g)

Answer 3

Câu 8:

nH2SO4=0,15.1=0,15(mol)

nNaOH=0,15.1=0,15(mol)

PTHH:  2 NaOH + H2SO4 -> Na2SO4 + 2 H2O

Ta có: 0,15/2 < 0,15/1

=> NaOH hết, H2SO4 dư => Dư axit => Qùy tím hóa đỏ

Answer 4

Câu 7:

nHCl= 0,16.1=0,16(mol)

PTHHl R +2 HCl -> RCl2 + H2

0,08______0,16(mol)

nR=1/2. nHCl=0,16/2=0,08(mol)

=>M(R)=mR/nR=5,2/0,08=65(g/mol)

=> R(II) cần tìm là Kẽm (Zn=65)