дано
m(ppa NaOH) = 200 g
W(NaOH) = 2%
m практ (Cu(OH)2) = 4 g
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η(Cu(OH)2)-?
m(NaOH) = 200 * 2% / 100% = 4 g
M(NaOH) = 40 g/mol
n(NaOH)= m/M =4/ 40 = 0.1 mol
2NaOH+CuSO4-->Cu(OH)2+Na2SO4
2n(NaOH) = n(Cu(OH)2
n(Cu(OH)2) = 0.1 / 2 = 0.05 mol
M(CuSO4) = 160 g/mol
m теор(CuSO4) = n*M =0.05 * 160 = 8 g
η(Cu(OH)2) = mпракт (Cu(OH)2) / m теор(Cu(OH)2) * 100% = 4/8 * 100% = 50%
ответ 50%