a)
nCuSO4.5H2O = 5/250 = 0.02 mol = n CuSO4 => mCuSO4 = 3.2g
m dd sau = mCuSO4.5H2O + mH2O = 5 + 19.5 = 24.5g
C % = mCuSO4 x 100%/mdd sau pư = 3.2 x 100%/24.5 = 13.06%
b)
nCuSO4.5H2O = 5/250 = 0.02 mol
nCuSO4 sau= nCuSO4.5H2O + ndd CuSO4 = 5/250 + 200 x 4/(100 x 160) = 0.07 mol
=> mCuSO4 = 0.07 x 160 = 11.2g
mdd = 5 + 200 = 205g
=> C % = 11.2 x 100% / 205 = 5.46%
c)
mNaOH = 8 + 200x10/100 = 20g
mdd = 8 + 200 = 208g
=> C% = 20 x 100%/208 = 9.62%
d)
Na2O + H2O -> 2NaOH
0.1 0.2
tổng mNaOH = 0.2x40 + 200x10/100 = 28g
mdd = 6.2 + 200 = 206.2g
=> C% = 28x100%/206.2 = 13.58%
e)
2Na + 2H2O -> 2NaOH + H2
0.2 0.2 0.1
mNaOH =0.2x40 + 200x10/100 = 28g
mdd = mNa + mdd NaOH - mH2 = 4.6 + 200 - 0.1 x 2 =204.4g
=> C % = 28/204.4 = 13.7%