Đáp án:

 

Giải thích các bước giải:

 $CuSO_4+BaCl_2→CuCl_2+BaSO_4$

Ta có
$m_{CuSO4}=\frac{500.4}{100}=20(g)$ 

⇒$n_{CuSO4}=20:160=0,125(mol)$

$n_{BaCl2}=300.5,2\%=15,6(g)$

⇒$n_{BaCl2}=\frac{15,6}{208}=0,075(mol)$

⇒CuSO4 dư

$n_{CuSO4}dư=0,125-0,075=0,05(mol)$

$m_{CuSO4}dư=0,05.160=8(g)$

$m_{BaSO4}=0,075.233=17,475(g)$

$m{dd sau pư}=500+300-17,475=782,525(g)$

$C\%_{CuSO4dư}=\frac{8}{782,525}.100\%=1.02\%$

$m_{CuCl2}=0,075.135=10,125(g)$

$C\%_{CuCl2}=\frac{10,125}{782,525}.100\%=1,29\%$

ta có:
mCuO4=500.4/100=20g
=>nCuSO4=20/160=0,125 mol
mBaCl2=300.5,2/100=15,6g
=>BaCl2=15,6/208=0,075 mol
PTHH:
CuSO4+BaCl2---->CuCl2+BaSO4
1________1
0,125____0,075
Theo pt,CuO4 dư,BaCl2 hết
=>dd B gồm CuO4 dư và CuCl2
=>nBaSO4=nBaCl=0,075 mol
=>mBaSO4=0,075.233=17,475 g
m dd B=500+300-17,475=782,525 g
theo PT=>nCuSO4 dư=0,125-0,075=0,05 mol
=>mCuSO4 dư=8 g
=>%mCuSO4=8.100%/782,525=1,02%
nCuCL2=0,075 mol
=>mCuCL2=10,125 g

=> %mCuCL2=10,125.100%/782,525=1,29%