a)2NaOH+CuSO4--->Cu(OH)2+Na2SO4
Vậy Alà Na2SO4
B là Cu(OH)2
b) Ta có
m\(_{NaOH}=\frac{100.20}{100}=20\left(g\right)\)
n NaOH=20/40=0,5(mol)
Theo pthh
n CuSO4=1/2n NaOH=0,25(mol)
m CuSO4=0,25.160=40(g)
x=mdd CuSO4=\(\frac{40.100}{16}=250\left(g\right)\)
c)
Theo pthh
n Cu(OH)2=1/2 n NaOH=0,25(mol)
m CuSO4=0,25.98=24,5(g)
n Na2SO4=1/2n NaOH=0,25(mol)
m na2SO4=0,25.142=35,5(g)
m dd=250+100-24,5=325,5(g)
C%=35,5/325,5.100%=10,9%