30cm^3 of a mixture of methane and ethyne require 70cm^3 of oxygen for complete combustion. Calculate the composition of the mixture
This is a problem with two unknowns (at least the way I'm solving it) so it requires two equations and you solve them simultaneously.
Write and balance the equations.
CH4 + 2O2 ==> CO2 + 2H2O
2C2H2 + 5O2 ==> 4CO2 + 2H2O
Let X = cc of CH4
and Y = cc of C2H2
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equation 1 is X + Y = 30 cc
equation 2 is cc oxygen in tems of X and Y
(2X) is cc O2 in terms of X for combustion of CH4 and
(5Y/2) is cc O2 in terms of Y for combustion of C2H2. Put those together to get equation 2 which is
2X + 5Y/2 = 70 cc..........equation 2
X + Y = 30 cc ..............equation 1
Solve for X and Y.
The problem doesn't say how it wants the composition; usually this is done as a percent so
(X/70)*100 = %CH4
(Y/70)*100 = %C2H2
Post your work if you get stuck.
To calculate the composition of the mixture, we need to determine the amount of methane and ethyne present.
Let's assume that the mixture contains x cm^3 of methane and y cm^3 of ethyne.
According to the given information, 30 cm^3 of the mixture requires 70 cm^3 of oxygen for complete combustion.
Since the combustion reaction involves the ratio of 1:1 between methane and oxygen, and 2:1 between ethyne and oxygen, we can set up the following equation based on the volume of oxygen used:
x + (2/3)y = 70 (equation 1)
Next, we can consider the volume of the mixture. Since the volume is 30 cm^3, we can set up the following equation based on the volume of the mixture:
x + y = 30 (equation 2)
Now we have a system of two equations (equation 1 and equation 2) that we can solve simultaneously.
Let's solve these equations using substitution method:
From equation 2, we can isolate x:
x = 30 - y
Substitute this value into equation 1:
(30 - y) + (2/3)y = 70
30 - y + (2/3)y = 70
(1/3)y = 40
y = (40 * 3) / 1
y = 120 cm^3
Substitute the value of y back into equation 2 to find x:
x + 120 = 30
x = 30 - 120
x = -90 cm^3
Since we cannot have a negative volume, this means that x = 0.
Therefore, the composition of the mixture is 0 cm^3 of methane and 120 cm^3 of ethyne.
To calculate the composition of the mixture, we need to understand the concept of stoichiometry. Stoichiometry is the calculation of reactants and products in chemical reactions based on their balanced equations.
In this case, we have a mixture of methane (CH4) and ethyne (C2H2), which reacts with oxygen (O2) to produce carbon dioxide (CO2) and water (H2O). The balanced equation for the combustion of methane is:
CH4 + 2O2 → CO2 + 2H2O
The balanced equation for the combustion of ethyne is:
2C2H2 + 5O2 → 4CO2 + 2H2O
Now, let's calculate the volume of oxygen required for the combustion of 30cm^3 of the mixture.
By comparing the coefficients in the balanced equations, we can determine that for every 1 cm^3 of methane, 2 cm^3 of oxygen is required, and for every 2 cm^3 of ethyne, 5 cm^3 of oxygen is required.
Let's assume x cm^3 of methane and y cm^3 of ethyne are present in the mixture.
Therefore, the total volume of methane in the mixture is 30 * (x / (x + y)) cm^3, and the total volume of ethyne is 30 * (y / (x + y)) cm^3.
According to the given information, 70 cm^3 of oxygen is required for complete combustion. It can be written as:
(2 * (x / (x + y))) + (5 * (y / (x + y))) = 70
Simplifying the equation, we get:
2x + 5y = 70 * (x + y)
Expanding it further:
2x + 5y = 70x + 70y
Rearranging the terms, we have:
68x - 65y = 0
To solve this system of equations, we need another equation. In this case, we can use the fact that the sum of the volumes of methane and ethyne should be equal to 30 cm^3:
x + y = 30
Now, we have a system of equations:
68x - 65y = 0 (Equation 1)
x + y = 30 (Equation 2)
To solve this, we can use any method, such as substitution or elimination. Here, we will use the substitution method.
Rearrange Equation 2 to get:
x = 30 - y
Substitute this value of x in Equation 1:
68(30 - y) - 65y = 0
Simplifying:
2040 - 68y - 65y = 0
Combining like terms:
-133y = -2040
Divide by -133 on both sides:
y ≈ 15.34 cm^3
Substitute this value of y back into Equation 2 to find x:
x + 15.34 ≈ 30
x ≈ 14.66 cm^3
Therefore, the composition of the mixture is approximately 14.66 cm^3 of methane and 15.34 cm^3 of ethyne.
Note: It is important to remember that this is an approximate calculation.