Hello, Emma,
I believe the equation for combustion of ethyne should have a coefficient of 2 for the H2O.
2C2H2 + 5O2 = 4CO2 + 2H2O
It turnws out, however, that we don't need the correct number for the water produced - just the CO2. The molar ratio of CO2 to C2H2 is 2/1. We'll get twice the moles of carbon dioxide as we have of the ethyne, assuming excess oxygen is present. So let's calculate how many moles we have in 84.4 grams of the C2H2.
Divide the mass of ethyne by it's molar mass to find moles C2H2. 84.4 grams/26.02 g/mole = 3.243 moles of C2H6. Since we expect twice that amount of CO2, the reaction will produce 2*3.24 moles = 6.486 moles of carbon dioxide. Please note that I did not round to the nearest hundredth on the initial calculation, so when the 3.243 is doubled, the 0.003 portion becomes 0.006, and so the when the result of 6.486 is rounded to the nearest hundreth, it results in 6.49 moles, not 6.48. I used the more accurate figure of 6.49 in calculation final volume based on the gas laws.
To find the volume of 6.49 moles of CO2 at the conditions noted, use the ideal gas law, PV = nRT. The units of R are L*atm/mol*K. Pressure is already in atm, but temperature must be converted from C to K. Add 273.15 to the temperature to yield 335.35K. Note that the volume unit is liters in this calculation, since R has the liter unit. We'll need to convert that to ml later.
Rearrange the ideal gas law to isolate the unknown, V.
V = nRT/P
Enter the data, and R, and calculate. I found 129.3 liters. Multiply by 1000 ml/liter to get ml. 129,300 ml.
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[extra credit next]
129,300ml seems like a lot, but remember, at STP, 1 mole of any gas occupies 22.4 liters. These conditions deviate from STP in that the pressure is higher than standard pressure of 1 atm, and the temperature is also higher than standard temperature of 0 C. The higher pressure should requce the volume, while the higher temperature will increase it.
Out of curiosity, let's calculate the value of 6.49 moles of gas at STP and then adjust it for the non-STP conditions.
At STP: 6.49 moles*(22.4 liters/mole) = 145.4 liters CO2. Adjust this for the new temperature and pressure using:
P1V1/T1 = P2V2/T2
The initial state, denoted by the subscript 1, are the STP conditions. The final state is marked with thew subscript 2, which are the conditions under which the gas is actually measured. I want to see if the calculated volume of gas at STP (145.4 liters) will equal the volume we calculated using the ideal gas law at the actual conditions.
Rearrange to find the unknown, V2:
V2 = V1(T2/T1)(P1/P2)
Convert temperatures into K and enter the data. Note how I've grouped the terms as ratios of the temperatures and pressures. This makes it easier to see what is happening. In going from STP to the actual conditions, both the temperature and pressure increase. The temperature ratio (T2/T1 or ) will therefore be greater than 1 and increase the final volume, while the pressure ratio (P1/P2) will be less than 1 and decrease the volume.
I found the final volume using this approach to be 129.3 liters (or 129,300 ml). That compares to the 0riginal calculation of 129.3 liters. The same, as we would expect. This method is equally valid in calculating the volume of a gas when we know the moles and final conditions. I've used it when I couldn't remember the value for R, and didn't feel like looking it up.
I hope this helps,
Bob
