The following is the balanced equation of the combustion of methane gas with oxygen gas:CH4(g) + 2O2(g) -> CO2(g) + 2H2O(l)If you ran a reaction in a sealed balloon at 25o C with 1 g of methane, CH4, and 3 g of oxygen, O2, what is the end volume of the balloon once the reaction is complete? The pressure was constant at 1 atm. Assume all gases behave ideally and you have 100% yield. (ignore vapor pressure of water)A. 4.9 LB. 0.096 LC. 1.1 LD 1.5 LE. 0.13 L
Question
The following is the balanced equation of the combustion of methane gas with oxygen gas:CH4(g) + 2O2(g) -> CO2(g) + 2H2O(l)If you ran a reaction in a sealed balloon at 25o C with 1 g of methane, CH4, and 3 g of oxygen, O2, what is the end volume of the balloon once the reaction is complete? The pressure was constant at 1 atm. Assume all gases behave ideally and you have 100% yield. (ignore vapor pressure of water)A. 4.9 LB. 0.096 LC. 1.1 LD 1.5 LE. 0.13 L
Solution
To solve this problem, we need to use the ideal gas law, which states that PV = nRT, where P is the pressure, V is the volume, n is the number of moles of gas, R is the ideal gas constant, and T is the temperature in Kelvin.
First, we need to convert the temperature from Celsius to Kelvin. The formula to convert Celsius to Kelvin is K = C + 273.15. So, 25°C = 298.15 K.
Next, we need to calculate the number of moles of each reactant. The molar mass of methane (CH4) is approximately 16.04 g/mol and the molar mass of oxygen (O2) is approximately 32 g/mol. So, we have:
1 g CH4 * (1 mol/16.04 g) = 0.062 mol CH4 3 g O2 * (1 mol/32 g) = 0.094 mol O2
The balanced chemical equation tells us that one mole of methane reacts with two moles of oxygen to produce one mole of carbon dioxide and two moles of water. However, we have less than one mole of each reactant. The reaction will stop when one of the reactants is used up. In this case, methane is the limiting reactant because we have less of it. So, the reaction will produce 0.062 mol of carbon dioxide.
Finally, we can use the ideal gas law to calculate the volume of carbon dioxide produced. We know that P = 1 atm, n = 0.062 mol, R = 0.0821 Latm/(molK), and T = 298.15 K. Solving for V gives us:
V = nRT/P V = (0.062 mol)(0.0821 Latm/(molK))(298.15 K) / 1 atm V = 1.5 L
So, the end volume of the balloon once the reaction is complete is 1.5 L. The correct answer is D. 1.5 L.
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