Задача номер 6, если не сложно то и номер 7

6)

дано

m(ppa CuSO4) = 35 g

W(CuSO4) = 20%

m( ppa NaOH) = 20 g

W(NaOH) = 25 %

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m(Cu(OH)2)-?

m(CuSO4) = m(ppa CuSO4) * W(CuSO4) / 100% = 35*20% / 100% = 7 g

M(CuSO4) =160 g/mol

n(CuSO4) = n/M = 7 / 160 = 0.04375 mol

m(NaOH) = m(ppa NaOH) * W(NaOH) / 100% = 20*25% / 100% = 5 g

M(NaOH) = 40 g/mol

n(NaOH) = n/M = 5/ 40 = 0.125 mol

n(CuSO4) < n(NaOH)

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

n(CuSO4) = Cu(OH)2 = 0.04375 mol

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

m(Cu(OH)2) = n*M = 0.04375 * 98 = 4.2875 g

ответ 4.2875

7)

дано

m техн(WO3) = 240 g

W(прим) = 5%

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V(H2)-?

m чист(WO3) = m техн(WO3) - (m техн (WO3) * W(прим) ) / 100%

m чист (WO3)= 240 - (240*5%) / 100% = 228 g

2WO3+3H2-->2W+3H2O

M(WO3) = 232 g/mol

n(WO3) = m/M = 228 / 232 = 0.98 mol

2n(WO3) = 3n(H2)

n(H2) = 3*0.98 / 2 = 1.47 mol

V(H2) = n*Vm = 1.47 * 22.4 = 32.928 L

ответ 32.928 л