A mixture of methane (CH4) and ethane (C2H6) of mass 13.43g is completely burned in oxygen.

If the total mass of CO2 and H2O produced is 64.84 g, calculate the fraction of CH4 in the mixture.

CH4 + C2H6 arrow CO2 + H2O
13.43g 64.84g

Would I do 13.43g/16.04g CH4 ?

I don't get this.

Start with the chemical equations of the complete combustion of CH4 and C2H6,

CH4 + mO2 -> CO2 + 2H2O ....(1)
C2H6 + nO2 -> 2CO2 + 3H2O ....(2)
Calculate the total mass of CO2+H2O produced if all 13.43g of gas were CH4.
Do the same for C2H6.
Find the ratio of each gas by proportion.

Would I solve for m

13.43 + m32 = 36.04

Also for CO2 + 2H2O ....(1)

2CO2 + 3H2O ....(2)

why does the bottom one have 2CO2 and 3H2O?

First, you don't need to know m and n, because only the products count. However, m=2 and n=3.5.

If written properly, equation 2 should be all multiplied by 2.

For equation 1, calculate the RMM for each ingredient, then calculate the ratio of masses of CH4 and the products, namely CO2 and 2H2O.
Thus (12+4)=16 g of CH4 will produce (12+32 + 2(2+16))=80 g of CO2 and H2O combined. The ratio of mass of products/mass of CH4 is therefore 80/16=5.

Calculate similarly the ratio of mass of products / mass of C2H6, call it x.

The ratio of mass of products / mass of mixture = 64.84/13.43 = 4.828

From the ratios 5 and x, calculate the mix of CH4 and C2H6 to give a ratio of 4.828.

The explanation is poor try putting up the answer ! So people can see the way to proably do it!

To determine the fraction of CH4 in the mixture, you need to calculate the mass of CH4 in the mixture and then divide it by the total mass of the mixture.

First, you need to calculate the molar mass (mass of one mole) of CH4 and C2H6. The molar mass of CH4 is calculated as follows:

1 carbon atom (C) * atomic mass of carbon (12.01 g/mol) + 4 hydrogen atoms (H) * atomic mass of hydrogen (1.01 g/mol) = 16.04 g/mol

Now, calculate the number of moles of CH4 in the mixture by dividing the mass of CH4 by its molar mass:

moles of CH4 = mass of CH4 / molar mass of CH4
moles of CH4 = 13.43 g / 16.04 g/mol ≈ 0.836 mol

Since CH4 and C2H6 react completely, the moles of CO2 and H2O produced will also be equal to the moles of CH4 and C2H6 in the mixture.

Next, calculate the mass of CO2 and H2O produced by adding their respective masses:

mass of CO2 + mass of H2O = 64.84 g

The molar masses of CO2 and H2O are calculated as follows:

CO2: 1 carbon atom (C) * atomic mass of carbon (12.01 g/mol) + 2 oxygen atoms (O) * atomic mass of oxygen (16.00 g/mol) = 44.01 g/mol

H2O: 2 hydrogen atoms (H) * atomic mass of hydrogen (1.01 g/mol) + 1 oxygen atom (O) * atomic mass of oxygen (16.00 g/mol) = 18.02 g/mol

Now, let's calculate the moles of CO2 and H2O produced:

moles of CO2 = mass of CO2 / molar mass of CO2
moles of CO2 = 44.01 g/mol

moles of H2O = mass of H2O / molar mass of H2O
moles of H2O = 18.02 g / 18.02 g/mol ≈ 3 mol

Since the moles of CO2 and H2O are equal to the moles of CH4 and C2H6, we can calculate the moles of C2H6:

moles of C2H6 = moles of CO2 - moles of CH4
moles of C2H6 = 3 mol - 0.836 mol ≈ 2.164 mol

Next, calculate the molar mass of C2H6:

2 carbon atoms (C) * atomic mass of carbon (12.01 g/mol) + 6 hydrogen atoms (H) * atomic mass of hydrogen (1.01 g/mol) = 30.07 g/mol

Now, calculate the mass of C2H6 in the mixture by multiplying the moles of C2H6 by its molar mass:

mass of C2H6 = moles of C2H6 * molar mass of C2H6
mass of C2H6 = 2.164 mol * 30.07 g/mol ≈ 65.05 g

Finally, calculate the fraction of CH4 in the mixture by dividing the mass of CH4 by the total mass of the mixture:

fraction of CH4 = mass of CH4 / (mass of CH4 + mass of C2H6)
fraction of CH4 = 13.43 g / (13.43 g + 65.05 g) ≈ 0.171

Therefore, the fraction of CH4 in the mixture is approximately 0.171, or 17.1%.