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.)