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