nBa=27,4:137=0,2(mol)
mCuSO4=500.2%=10(g)
nCuSO4=10:160=0,0625(mol)
PTHH:
Ba + 2H2O-->Ba(OH)2+H2 (1)
0,2 -> 0,2 ->0,2 ->0,2
Ba(OH)2+CuSO4-->BaSO4\(\downarrow\)+Cu(OH)2\(\downarrow\) (2)
0,2 0,0625
Theo PT(2) : ta có :Ba(OH)2 dư ; CuSO4 hết
=>nBa(OH)2 p/ứ = nBaSO4 = nCu(OH)2=0,0625(Mol)
nBa(OH)2 dư = 0,2-0,0625=0,1375(mol)
=>mBa(OH)2 dư =0,1375 . 171=23,5125(g)
mddsaup/ứ =500+0,2.137-0,2.2-0,0625.233-0,0625.98=506,3125(g)
%mBa(OH)2 dư =\(\frac{23,5125}{506,3125}.100\%\approx4,64\%\)