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Autok2015
@Autok2015
July 2022
1
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Срочно Определите простейшую формулу вещества, содержащего только углерод и водород, если при полном сгорании 2 г этого вещества образуется 4, 5 г воды.
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Alexei78
Дано
m(CxHy)=2 g
m(H2O) = 4.5 g
---------------------
CxHy-?
CxHy+O2--> CO2+H2O
M(H2O) = 18 g/mol
n(H2O) =m/M = 4.5 / 18 = 0.25 mol
n(H) = 2*n(H2O) = 2*0.25 = 0.5 mol
M(H)= 1 g/mol
m(H)= n(H)*M(H)= 1*0.5 = 0.5 g
m(C)=m(CxHy)-m(H) = 2-0.5 = 1.5 g
M(C) = 12 g/mol
n(C) = m/M = 1.5 / 12 = 0.125 mol
CxHy = 0.125 : 0.5 =1:4
CH4
ответ МЕТАН
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Answers & Comments
m(CxHy)=2 g
m(H2O) = 4.5 g
---------------------
CxHy-?
CxHy+O2--> CO2+H2O
M(H2O) = 18 g/mol
n(H2O) =m/M = 4.5 / 18 = 0.25 mol
n(H) = 2*n(H2O) = 2*0.25 = 0.5 mol
M(H)= 1 g/mol
m(H)= n(H)*M(H)= 1*0.5 = 0.5 g
m(C)=m(CxHy)-m(H) = 2-0.5 = 1.5 g
M(C) = 12 g/mol
n(C) = m/M = 1.5 / 12 = 0.125 mol
CxHy = 0.125 : 0.5 =1:4
CH4
ответ МЕТАН