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 ?
This a two equation problem that you solve simultaneously. Here are the two equations.
Let x = g CH4
and y = g C2H6
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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.
2.96
To determine the grams of methane in the original 5.00 g mixture, we need to use the given information about the production of carbon dioxide.
1. First, let's find the number of moles of CO2 produced. We can use the molar mass of CO2 to convert from grams to moles:
Molar mass of CO2 = 12.01 g/mol (C) + 2 * 16.00 g/mol (O) = 44.01 g/mol
Number of moles of CO2 produced = mass of CO2 / molar mass of CO2
= 14.09 g / 44.01 g/mol
= 0.3198 mol
2. Now, let's determine the number of moles of methane (CH4) that were present in the mixture. By comparing the chemical equation, we can see that the ratio of the moles of CO2 to CH4 is 1:1.
Number of moles of CH4 = number of moles of CO2 produced
= 0.3198 mol
3. Finally, let's find the mass of methane using its molar mass:
Molar mass of CH4 = 12.01 g/mol (C) + 4 * 1.008 g/mol (H) = 16.04 g/mol
Mass of CH4 = number of moles of CH4 * molar mass of CH4
= 0.3198 mol * 16.04 g/mol
= 5.12 g
Therefore, there were 5.12 grams of methane in the original 5.00 g mixture.
To solve this problem, we need to use the balanced chemical equation for the combustion of methane (CH4). The balanced equation is:
CH4 + 2O2 → CO2 + 2H2O
From the equation, we can determine that for every 1 mole of CH4 combusted, 1 mole of CO2 is produced.
First, calculate the number of moles of CO2 produced using its molar mass:
Molar mass of CO2 = 12.01 g/mol (C) + 2 * 16.00 g/mol (O) = 44.01 g/mol
Number of moles of CO2 produced = mass of CO2 / molar mass of CO2
Number of moles of CO2 produced = 14.09 g / 44.01 g/mol = 0.320 mol
Since 1 mole of CH4 produces 1 mole of CO2, the number of moles of CH4 present in the original 5.00 g mixture is also 0.320 mol.
Now, let's calculate the mass of CH4 using its molar mass:
Molar mass of CH4 = 12.01 g/mol (C) + 4 * 1.01 g/mol (H) = 16.05 g/mol
Mass of CH4 = number of moles of CH4 * molar mass of CH4
Mass of CH4 = 0.320 mol * 16.05 g/mol = 5.14 g
Therefore, there were approximately 5.14 grams of methane in the original 5.00 g mixture.