Question

at 120 C and 1 atm pressure, 1.00 L of methane, CH4, reacts completely with excess oxygen to from carbon dioxide and water. What volumes of the two products are produced at this pressure and temperature?

Answer 1

The volume of carbon dioxide produced is 1.91 L and the volume of water produced is 3.82 L.

Step-by-step explanation:

To determine the volumes of carbon dioxide and water produced in the reaction, we need to use the molar ratio of the reactants and products. The balanced equation for the reaction is:

CH4(g) + 2O2(g) → CO2(g) + 2H2O(g)

From the equation, we can see that 1 mole of methane reacts with 2 moles of oxygen to produce 1 mole of carbon dioxide and 2 moles of water. Since we know the initial volume of methane is 1.00 L, we can use the ideal gas law to calculate the number of moles of methane:

n = PV/RT

n = (1.00 atm)(1.00 L) / (0.0821 atm·L/mol·K)(120 + 273.15 K)

n = 0.0457 mol

According to the stoichiometry of the reaction, 0.0457 mol of methane will produce 0.0457 mol of carbon dioxide and 0.0914 mol of water. Using the ideal gas law again, we can calculate the volume of each product:

V = nRT/P

V(CO2) = (0.0457 mol)(0.0821 atm·L/mol·K)(120 + 273.15 K) / 1.00 atm

V(CO2) = 1.91 L

V(H2O) = (0.0914 mol)(0.0821 atm·L/mol·K)(120 + 273.15 K) / 1.00 atm

V(H2O) = 3.82 L