Question

36 g of water was produced when a volume of methane was burned in excess oxygen. assuming the reaction began at room temperature and pressure, what was the volume of the methane? use the information below to calculate your answer.

Answer 1

To calculate the volume of methane, we need to use the balanced chemical equation for the reaction between methane and oxygen. From the equation, we can see that for every mole of methane, 2 moles of water are produced. Therefore, the volume of methane is equal to the molar volume at STP.

Step-by-step explanation:

To calculate the volume of methane, we need to use the balanced chemical equation for the reaction between methane and oxygen:

CH4 + 2O2 -> CO2 + 2H2O

From the equation, we can see that for every mole of methane, 2 moles of water are produced. Since we know that 36 g of water was produced, we can calculate the number of moles of water: 36 g / 18.02 g/mol = 2 moles of water

Since 2 moles of water are produced from 1 mole of methane, we can conclude that 1 mole of methane was burned. Therefore, the volume of methane is equal to the molar volume at STP, which is approximately 22.414 L/mol.

Answer 2

The volume of methane burned to produce 36 g of water is 22.414 liters, derived from using the stoichiometric relationship in the balanced chemical equation for the combustion of methane and the known molar volume of a gas at STP.

Step-by-step explanation:

To determine the volume of methane that was burned to produce 36 g of water, we must first understand the stoichiometry of the reaction between methane and oxygen. The balanced chemical equation for the combustion of methane is:

CH4(g) + 2 O2(g) → CO2(g) + 2 H2O(l)

From the equation, we can see that 2 moles of water are produced for every mole of methane burned. Since the molar mass of water is approximately 18.02 g/mol, we can calculate the moles of water produced by dividing 36 g of water by 18.02 g/mol, giving us 2 moles of water. This means 1 mole of methane (16.04 g/mol) was burned, as the molar ratio of methane to water in the reaction is 1:2.

At standard temperature and pressure (STP), one mole of any gas occupies 22.414 liters. Therefore, the volume of methane burned to produce 36 g of water is 22.414 liters, since 1 mole of methane was burned.