a 5 g mixture of natural gas containing ch4 and c2h4 was burnt in excess of oxygen yielding 14.5 g of co2 and h2o.What is the weight of ch4 and co2 in the mixture
Am so sorry but I really not understand this 🙏
To find the weight of CH4 and CO2 in the mixture, we need to calculate the amount of each compound produced.
First, let's start by calculating the amount of CO2 produced.
- The given weight of CO2 produced is 14.5 g.
- The molar mass of CO2 is 44 g/mol (12 g/mol for each carbon atom and 16 g/mol for each oxygen atom).
Using the formula:
Amount (in mol) = Weight (in g) / Molar mass (in g/mol)
Amount of CO2 = 14.5 g / 44 g/mol
Amount of CO2 = 0.3295 mol
Now, let's calculate the amount of CH4 in the mixture.
We know that the initial weight of the mixture is 5 g, but we don't have the individual amounts of CH4 and C2H4.
However, we can assume that all the carbon in the mixture comes from CH4 since it is completely converted to CO2. Therefore, the amount of carbon in CO2 will be the same as the amount of carbon in CH4.
- The molar mass of CH4 is 16 g/mol (12 g/mol for carbon and 4 g/mol for hydrogen).
- CH4 contains one carbon atom, so the molar mass of carbon in CH4 is 12 g/mol.
Using the formula:
Amount of CH4 = Amount of carbon in CO2 = Amount of CO2 * (Molar mass of carbon in CH4 / Molar mass of CO2)
Amount of CH4 = 0.3295 mol * (12 g/mol / 44 g/mol)
Amount of CH4 = 0.0897 mol
To find the weight of CH4, we multiply the amount of CH4 by its molar mass:
Weight of CH4 = Amount of CH4 * Molar mass of CH4
Weight of CH4 = 0.0897 mol * 16 g/mol
Weight of CH4 = 1.435 g
Therefore, the weight of CH4 in the mixture is approximately 1.435 g, and the weight of CO2 is 14.5 g as given.
I am unable to solve the equations. Can you put up the equations too. The answer is a bit weird.
CH4 + 2O2 ==> CO2 + 2H2O
C2H4 + 3O2 ==> 2CO2 + 2H2O
Let x = mass CH4
and y = mass C2H4
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You need two equations.
(eqn 1) is total mass = 5g
(eqn 2) is (mass CO2+H2O from CH4) + (mass CO2+H2O from C2H4) = 14.5g
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Now convert those into x and y and solve. I have let mm stand for molar mass below.
(eqn 1) is x + y = 5
(eqn 2) is
[x(mmCO2/mmCH4)+x(2*mmH2O/mmCH4)]+y(2*mmCO2/mmC2H4)+y(2mmH2O)/mmC2H4] = 14.5g
Solve for x = mass CH4
y = mass C2H4.
Post your work if you get stuck.