G:
Gọi x,y là số mol CuO,Fe2O3
Theo gt:m(hh)=m(CuO)+m(Fe2O3)=80x+160y=40(1)
*Hòa tan hh vào dd HCl:
n(HCl)=0,7.2=1,4(mol)
mdd(HCl)=700.1,2=840(g)
Ta có PTHH:
CuO+2HCl->CuCl2+H2O(1)
x..........2x........x......................(mol)
Fe2O3+6HCl->2FeCl3+3H2O(2)
y................6y.......2y...................(mol)
Theo PTHH(1);(2):n(HCl)=2x+6y=1,4(mol)(2)
Giải PT(1);(2)=>x=0,1;y=0,2(mol)
=>n(CuCl2)=x=0,1(mol)=>m(CuCl2)=0,1.135=13,5(g)
=>n(FeCl3)=2y=2.0,2=0,4(mol)==>m(FeCl3)=0,4.162,5=65(g)
Ta có:mdd(sau)=m(hh)+mdd(HCl)=840+40=880(g)
=>C%CuCl2=13,5/880.100%=1,5%
=>C%FeCl3=65/880.100%=7,4%
b)% theo m các oxit:
m(CuO)=80x=80.0,1=8(g)
m(Fe2O3)=160y=160.0,2=32(g)
=>%m(CuO)=8/40.100%=20%
=>%m(Fe2O3)=32/40.100%=80%