1.

a)

Đặt n_{CuO}=x(mol);n_{Fe_2O_3}=y(mol)

->80x+160y=48(1)

800ml=0,8l

CuO+H_2SO_4->CuSO_4+H_2O

Fe_2O_3+3H_2SO_4->Fe_2(SO_4)_3+3H_2O

Theo PT: n_{H_2SO_4}=x+3y=0,8.1=0,8(2)

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

->m_{CuO}=0,2.80=16(g)

      m_{Fe_2O_3}=0,2.160=32(g)

b)

Theo PT: n_{CuSO_4}=x=0,2(mol);n_{Fe_2(SO_4)_3}=y=0,2(mol)

->C_{M\ CuSO_4}=C_{M\ Fe_2(SO_4)_3}={0,2}/{0,8}=4M

2.

a)

Đặt n_{CuO}=x(mol);n_{Fe_2O_3}=y(mol)

->80x+160y=3,2(1)

CuO+H_2  $$\xrightarrow{t^o}$$  Cu+H_2O

Fe_2O_3+3H_2  $$\xrightarrow{t^o}$$  2Fe+3H_2O

Theo PT: n_{H_2}=x+3y={1,2395}/{24,79}=0,05(2)

(1)(2)->x=0,02(mol);y=0,01(mol)

->\%m_{CuO}={0,02.80}/{3,2}.100\%=50\%

->\%m_{Fe_2O_3}=100-50=50\%

b)

Theo PT: n_{Cu}=x=0,02(mol);n_{Fe}=2y=0,02(mol)

->m_{Cu}=0,02.64=1,28(g)

       m_{Fe}=0,02.56=1,12(g)

Bài 1:

a)

Gọi n_{CuO} = a(mol); n_{Fe_2O_3} = b(mol)

=> 80a + 160b = 48(1)

n_{H_2SO_4} = 0,8.1 = 0,8(mol)

CuO + H_2SO_4 -> CuSO_4 + H_2O

a -> a -> a

Fe_2O_3 + 3H_2SO_4 -> Fe_2(SO_4)_3 + 3H_2O

b -> 3b -> b

=> a + 3b = 0,8(2)

(1),(2) => a = b = 0,2

\%m_{CuO} = (0,2.80)/(48) .100\% = 33,33\%

\%m_{Fe_2O_3} = 100\% - 33,33\% = 66,67\%

b)

C_{M(CuSO_4)} = C_{M(Fe_2(SO_4)_3)} = (0,2)/(0,8) = 0,25M

Bài 2:

Gọi n_{CuO} = a(mol); n_{Fe_2O_3} = b(mol)

=> 80a + 160b = 3,2(1)

n_{H_2} = (1,2395)/(24,79) = 0,05(mol)

$$CuO + H_2 \xrightarrow{t^o} Cu + H_2O$$

a -> a -> a

$$Fe_2O_3 + 3H_2 \xrightarrow{t^o} 2Fe + 3H_2O$$

b -> 3b -> b

=> a + 3b = 0,05(2)

(1),(2) => a = 0,02; b = 0,01

\%m_{CuO} = (0,02.80)/(3,2) .100\%= 50\%

\%m_{Fe_2O_3} = 100\% - 50\% = 50\%

b)

m_{Cu} = 0,02.64 = 1,28(g)

m_{Fe} = 0,01.2.56 = 1,12(g)