срочно задача на выход продукта даю 30 баллов

Ответ:

дано

m(ppa NaOH) = 120 g

W(NaOH) = 10%

m(CuSO4) = 16 g

m пр (Cu(OH)2) = 10 g

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η(Cu(OH)2)-?

m(NaOH) = 120 * 10% / 100% = 12 g

M(NaOH) = 40 g/mol

n(NaOH) = m/M = 12 / 40 = 0.3 mol

M(CuSO4) = 160 g/mol

n(CuSO4) = m/M = 16 / 160 = 0.1 mol

n(NaOH) > n(CuSO4)

2NaOH+CuSO4-->Na2SO4+Cu(OH)2

n(CuSO4) = n(Cu(OH)2) = 0.1 mol

M(Cu(OH)2) = 98 g/mol

m теор (Cu(OH)2) = n*M = 0.1 * 98 = 9.8 g

η(Cu(OH)2) = m теор (Cu(OH)2) / m пр (Cu(OH)2) * 100%

η(Cu(OH)2) = 9.8 / 10 * 100% = 98%

ответ 98%

Объяснение: