A 0.732-g mixture of CH4 and C2H6, is burned, yielding 2.059 g CO2. What is % of CH4 by mass in mixture?

Answer 1

Hi, Good Day!

I am not too sure if my solution is correct, I hope that it is useful for you.

Stoichiometric equation for both compounds respectively are:

CH4 + 2 O2 ------>CO2 + 2 H2O.....................(1)

C2H6 + 3.5 O2 -------> 2 CO2 + 3 H2O...........(2)

Let's assume weight of CH4 = X g
Then, weight of C2H6= (0.732-X) g

(Hint: Total mixture weight is 0.732 g)

Therefore, from Eqn (1) and Eqn (2) respectively:

1 mole of CH4 yields 1 mole of CO2 and
1 mole of C2H6 yields 2 moles of CO2

Thus, for CH4 : X/16 will yield X/16 moles of CO2
And, for C2H6 : (0.732-X)/30 will yield 2*(0.732-X)/30 moles of CO2.

(Hint: No. of moles = mass/molecular weight, MW of CH4=16, MW of C2H6=30)

Given that the weight of CO2 produced is 2.059g. Therefore,

Total moles of CO2 produced = 2.059/44

X/16 + 2*(0.732-X)/30 = 2.059/44

X = 0.4812 g

Therefore,

% of CH4 in mixture = (0.4812/0.732)*100%
= 65.74% by weight

Answer 2

Set up two equations in two unknowns, representing the number of moles of methane and ethane which could be present. The data supplied is sufficient to do this, and the result is obtained by algebra.

Answer 3

Do yer own homework.

Answer 4

srry not enough info can't help