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0,2 x
H2SO4+K2CO3=K2SO4+CO2+H2O
1 1
m(K2CO3)=m(р-ра)хω=110х0,25=27,5 г.
n(K2CO3)=m/M=27,5÷138=0,2 моль
n(CO2)=x=0,2х1÷1=0,2моль
V(CO2)=Vmхn=22,4л./моль×0,2моль=4,48л.
ответ: 4,48л.
Дано:
m(ClCH2-CH2-CH2-CH2-CH2Cl) = 286 г
m(пр. С5Н10) = 120 г
Найти:
η(С5Н10)-?
Решение.
М(ClCH2-CH2-CH2-CH2-CH2Cl) = 141 г/моль
n(ClCH2-CH2-CH2-CH2-CH2Cl) = m/M = 286 г/141 г/моль = 2,028 моль
M(C5H10) = 70 г/моль
n(пр. C5H10) = m/M = 120 г/70 г/моль = 1,714 моль
ClCH2-CH2-CH2-CH2-CH2Cl + 2Na = C5H10 + 2NaCl
Из УХР видно, что теоретически из 1 моль 1,5-дихлорпентана можно получить 1 моль циклопентана, т.е. применительно к данной задаче из 2,028 моль 1,5-дихлорпентана можно получить теоретически 2,028 моль циклопентана
η(С5Н10) = n(пр. С5Н10)/n(теор. С5Н10)*100% = 1,714/2,028*100% = 84,52%
ответ: 84,52%