Molar volume of an ideal gas at S.A.T.P. (25oC and 1.00 bar is 24.8 dm3 mol-1. Assume that gases are at 225oC and P=1.00 atmosphere before and after the combustion and behave as ideal gases.
The chemical equation for the combustion of butane is:
2C4H10(g) + 13O2(g) → 8CO2(g) + 10H2O(l)
At 225oC, we ignite a mixture of 5.00 dm3 of butane and 75.0 dm3 of O2
i. Is there enough oxygen for the complete combustion of the butane present?
ii. What gases are present in the resulting mixture after combustion?
iii. Calculate the final volume of the mixture after combustion (3 sig. fig).
Iv. Calculate the amounts (number of moles) of O2 and butane consumed using the molar volume of an ideal gas at S.A.T.P (3 sig. fig).
v.The final mixture is cooled down from 225oC to 25oC. Calculate the volume of the resulting gaseous mixture (3 sig. fig.).
There is a trick on using SATP. Note that one bar is not one atmosphere, If I recall, it is .986atm. Check that.
Secondly, the degree symbol (little superscripted zero) is not used with C or K, you just write 25C, or 300K, and when read, you do NOT say degrees. Twenty-five celcius is correct.
The problem is when you write 225oC, that is either 225C ,or 2250C, which are plenty different.
Now on your problem, calculate the moles of butane and O2 first. Look at the mole ratio in the balanced equation, which is deficient in quantity? After the reaction, you have along with the products, the reactant which was in excess?
To answer these questions, let's break down each part step by step.
i. To determine if there is enough oxygen for the complete combustion of butane, we need to use the stoichiometry of the balanced chemical equation.
From the balanced equation:
2C4H10(g) + 13O2(g) -> 8CO2(g) + 10H2O(l)
We can see that for every 2 moles of butane (C4H10), we need 13 moles of oxygen (O2). Therefore, for the complete combustion of 5.00 dm3 of butane, we would need:
(5.00 dm3 butane) x (13 mol O2 / 2 mol butane) = 32.5 dm3 of oxygen.
Since there are only 75.0 dm3 of oxygen available, which is greater than 32.5 dm3 required, there is enough oxygen for complete combustion.
ii. The resulting gases after combustion are carbon dioxide (CO2) and water vapor (H2O). These are the products of the combustion reaction as given by the balanced equation.
iii. To calculate the final volume of the mixture after combustion, we need to consider the volumes of the gases involved.
The initial volume of the mixture is 5.00 dm3 of butane + 75.0 dm3 of oxygen = 80.0 dm3.
After combustion, the water vapor (H2O) condenses to liquid form, so its volume can be neglected. Thus, the final volume of the resulting mixture would be the volume of carbon dioxide (CO2) produced.
From the balanced equation, we know that for every 2 moles of butane, we produce 8 moles of carbon dioxide. Using the molar volume of an ideal gas at S.A.T.P (24.8 dm3/mol), we can calculate the final volume of the mixture.
(5.00 dm3 butane) x (8 mol CO2 / 2 mol butane) x (24.8 dm3/mol) = 99.2 dm3.
Therefore, the final volume of the mixture after combustion is 99.2 dm3.
iv. To calculate the amounts (number of moles) of oxygen (O2) and butane consumed, we can use the molar volume of an ideal gas at S.A.T.P.
From the volume of oxygen needed for complete combustion calculation in part i, we found that 32.5 dm3 of oxygen is required.
Using the molar volume of an ideal gas at S.A.T.P (24.8 dm3/mol), we can calculate the number of moles consumed:
(32.5 dm3 O2) / (24.8 dm3/mol) = 1.31 mol O2 (rounded to 3 sig. fig.)
Similarly, for the butane (C4H10) consumed:
(5.00 dm3 butane) / (24.8 dm3/mol) = 0.202 mol butane (rounded to 3 sig. fig.)
v. To calculate the volume of the resulting gaseous mixture after cooling down to 25oC, we can use the ideal gas law equation:
PV = nRT
Where:
P = pressure (1.00 atmosphere)
V = volume (unknown)
n = number of moles (sum of moles of consumed gases from part iv.)
R = ideal gas constant (0.0821 L·atm/(mol·K))
T = temperature in Kelvin (25°C = 298 K)
Rearranging the equation to solve for V:
V = (nRT) / P
Calculating the final volume:
V = [(1.31 mol O2 + 0.202 mol butane) x (0.0821 L·atm/(mol·K)) x (298 K)] / (1.00 atm) = 34.15 L
Therefore, the volume of the resulting gaseous mixture after cooling down to 25°C is 34.15 L (rounded to 3 sig. fig.)