Asked by Ana
A 5.00 g mixture of methane (CH4) and ethane (C2H6) is combusted in oxygen gas to produce carbon dioxide and water. If 14.09 g of CO2 is produced, how many grams of methane was in the original 5.00 g mixture ?
Answers
Answered by
DrBob222
This a two equation problem that you solve simultaneously. Here are the two equations.
Let x = g CH4
and y = g C2H6
----------------
equation 1 is x + y = 5.00 g
You get the second equation by converting grams of CH4 to grams CO2 and grams C2H6 to grams CO2 and make the total add to 14.09 g CO2.
The chemical equations are
CH4 + 2O2 ==> CO2 + 2H2O
2C2H6 + 7O2 ==> 4CO2 + 6H2O
Convert g CH4 to g CO2.
x(mm CO2/mm CH4) where mm stands sfor molar mass.
Convert g C2H6 to g CO2
y(4*mm CO2/2*mmC2H6)
The total is 14.09g so put all of that together for equation 2 of
x(mm CO2/mm CH4) + y(4*mm CO2/2*mm C2H6) -= 14.09.
Now solve the two equations simultaneously and solve for x = grams CH4.
Let x = g CH4
and y = g C2H6
----------------
equation 1 is x + y = 5.00 g
You get the second equation by converting grams of CH4 to grams CO2 and grams C2H6 to grams CO2 and make the total add to 14.09 g CO2.
The chemical equations are
CH4 + 2O2 ==> CO2 + 2H2O
2C2H6 + 7O2 ==> 4CO2 + 6H2O
Convert g CH4 to g CO2.
x(mm CO2/mm CH4) where mm stands sfor molar mass.
Convert g C2H6 to g CO2
y(4*mm CO2/2*mmC2H6)
The total is 14.09g so put all of that together for equation 2 of
x(mm CO2/mm CH4) + y(4*mm CO2/2*mm C2H6) -= 14.09.
Now solve the two equations simultaneously and solve for x = grams CH4.
Answered by
Jonelle
2.96
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