6.

m_{dd\ H_2SO_4}={4,9}/{40\%}=12,25(g)

->V_{dd\ H_2SO_4}={12,25}/{1,31}\approx 9,35(ml)

7.

a)

CuO+2HCl->CuCl_2+H_2O

ZnO+2HCl->ZnCl_2+H_2O

b)

n_{CuO}=x(mol);n_{ZnO}=y(mol)

->80x+81y=12,1(1)

Theo PT: 2x+2y=n_{HCl}=0,1.3=0,3(2)

(1)(2)->x=0,05;y=0,1

->\%m_{CuO}={0,05.80}/{12,1}.100\%\approx 33,06\%

->\%m_{ZnO}=100-33,06=66,94\%

8.

n_{Na_2CO_3}=n_{Na_2CO_3.10H_2O}={28,6}/{286}=0,1(mol)

200ml=0,2l

->C_{M\ Na_2CO_3}={0,1}/{0,2}=0,5M

9.

n_{HCl}=0,2.0,5=0,1(mol)

Fe_2O_3+6HCl->2FeCl_3+3H_2O

MgO+2HCl->MgCl_2+H_2O

CuO+2HCl->CuCl_2+H_2O

Theo PT: n_{H_2O}=1/2n_{HCl}=0,05(mol)

BTKL: 2,8+0,1.36,5=0,05.18+m

->m=5,55(g)

10.

n_{KOH}={200.5,6\%}/{56}=0,2(mol)

2KOH+CuCl_2->Cu(OH)_2+2KCl

Theo PT: n_{Cu(OH)_2}=1/2n_{KOH}=0,1(mol)

->m_{Cu(OH)_2}=0,1.98=9,8(g)

11.

Có C\%={m_{NaCl}}/{m_{NaCl}+200}.100=32

->m_{NaCl}\approx 94,12(g)

Đáp án:

Giải thích các bước giải:

 6) số ml dd H2SO4 cần dùng là m= D.V

-> V= $$\frac{m}{D}$$ = $$\frac{4,9}{1,31}$$ = 3,74 ml 

7)