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nastenka60
@nastenka60
December 2021
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При спалюванні вуглеводню масою 4.2 г утворилось 13.2 г карбон (IV) оксиду і 5.4 г води. Відносна густина парів вуглеводню за воднем - 42. Знайдіть молекулярну формулу вуглеводню.
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Alexei78
Verified answer
Дано
m(CxHy)=4.2 g
m(CO2)=13.2g
m(H2O)=5.4g
D(H2)=42
----------------
CxHy-?
0.05 0.3 0.3
CxHy+O2-->CO2+H2O
1 mol x mol 0.5y
Mr(CxHy)=42*2=84
n(CxHy)=4.2/84=0.05 mol
n(CO2)=13.2/44=0.3 mol
x=0.3*1/0.05=6 - (C)
m(C)=0.3*12=3.6 g
m(H)=4.2-3.6=0.6 g
n(H)=0.6/1=0.6 mol
n(H2O)=0.6*0.5=0.3 mol
0.3*0.5=0.15
y=6*0.3/0.15=12 - (H)
CxHy=6:12
C6H12
ответ гексен
38 votes
Thanks 59
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Answers & Comments
Verified answer
Даноm(CxHy)=4.2 g
m(CO2)=13.2g
m(H2O)=5.4g
D(H2)=42
----------------
CxHy-?
0.05 0.3 0.3
CxHy+O2-->CO2+H2O
1 mol x mol 0.5y
Mr(CxHy)=42*2=84
n(CxHy)=4.2/84=0.05 mol
n(CO2)=13.2/44=0.3 mol
x=0.3*1/0.05=6 - (C)
m(C)=0.3*12=3.6 g
m(H)=4.2-3.6=0.6 g
n(H)=0.6/1=0.6 mol
n(H2O)=0.6*0.5=0.3 mol
0.3*0.5=0.15
y=6*0.3/0.15=12 - (H)
CxHy=6:12
C6H12
ответ гексен