Question

Помогите решить две задачи по органической химии

Answer 1

дано

V прим(C6H6) = 19.5 mL

η(C6H6) =80%

W(CaC2) = 96%

p(C6H6) = 0.8 g/cm3

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m тех(CaC2)-?

CaC2+2H2O-->Ca(OH)2+C2H2

3C2H2--->C6H6

m(C6H6) = p(C6H6) * V(C6H6) = 0.8 * 19.5 = 15.6 g

M(C6H6) = 78 g/mol

n(C6H6) = m/M =15.6 / 78 = 0.2 mol

3n(C2H2) = n(C6H6)

n(C2H2) = 3*0.2 = 0.6 mol

M(C2H2) = 26 g/mol

m прак (C2H2) = n*M = 0.6 * 26 = 15.6 g

m(теор C2H2) = 15.6 * 100% / 80% = 19.5 g

n(C2H2) m/M = 19.5 / 26 = 0.75 mol

n(CaC2) = n(C2H2) = 0.75 mol

n(CaC2) = 0.75 mol

M(CaC2) = 64 g/mol

m чист (CaC2) = n*M = 0.75 * 64 = 48 g

m техн(CaC2)  = 48 * 100% / 96% = 50 g

ответ 50 г

2)

дано

m практ (C6H6) = 1000 g

η(C6H6) = 82%

η(C2H2) = 10%

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V(CH4)-?

2CH4-->C2H2+3H2

3C2H2-->C6H6

m(теор C6H6) = 1000 * 100% / 82% = 1219.5 g

M(C6H6) = 78 g/mol

n(C6H6) = m/M = 1219.5 / 78 = 15.63 mol

3n(C2H2) = n(C6H6)

n(C2H2) = 15.63 * 3 = 46.95 mol

m теор (C2H2) = 46.95 * 100% / 10% = 469.5 g

M(C2H2) = 26 g/mol

n(C2H2) = m/M = 469.5 / 26 = 18.06 mol

2n(CH4) = n(C2H2)

n(CH4) = 2*18.06 = 36.12 mol

V(CH4) = n*Vm = 36.12 * 22.4= 809 L

ответ 809 л