Answer:
The balanced chemical equation for the reaction between methane and oxygen is:
CH4 + 2O2 → CO2 + 2H2O
From the equation, we can see that one mole of methane reacts with two moles of oxygen to produce one mole of carbon dioxide and two moles of water. We can use this information to determine the number of moles of methane and oxygen that react in the given volume of gases.
At STP (Standard Temperature and Pressure), one mole of any gas occupies a volume of 22.4 L. Therefore, the number of moles of methane present in 25 cm³ at STP is:
n_CH4 = V_CH4 / V_molar = (25/1000) / 22.4 = 0.00111607 mol
where V_molar is the molar volume of gas at STP.
Similarly, the number of moles of oxygen present in 65 cm³ at STP is:
n_O2 = V_O2 / V_molar = (65/1000) / 22.4 = 0.00290179 mol
The balanced chemical equation shows that methane and oxygen react in a 1:2 ratio. Therefore, all of the methane will be consumed by 0.00111607 mol of oxygen, leaving 0.00290179 - 0.00111607 = 0.00178572 mol of oxygen unreacted.
To find the volume of the residual gases, we can use the ideal gas law:
PV = nRT
where P is the pressure (which is 1 atm at STP), V is the volume, n is the number of moles, R is the gas constant (0.0821 L·atm/mol·K), and T is the temperature (273 K at STP).
Solving for V, the volume of the residual gases is:
V = nRT/P = (0.00178572 + x) * 0.0821 * 273 / 1
where x is the number of moles of other gases present in the residual mixture.
Since the volume of the residual gases at STP must be equal to the starting volume of the gases, we can set up an equation:
V_CH4 + V_O2 = V_residual
Substituting the given values, we have:
25/1000 + 65/1000 = (0.00178572 + x) * 0.0821 * 273 / 1
Solving for x, we get:
x = 0.00223675 mol
Therefore, the total number of moles of gas in the residual mixture is 0.00178572 + 0.00223675 = 0.00402247 mol. Substituting into the ideal gas law equation, we get:
V = nRT/P = 0.00402247 * 0.0821 * 273 / 1 = 0.0915 L
So the volume of the residual gases after sparking is approximately 0.0915 L at STP.