Question

How many moles of methane gas will combust to produce 330 grams of water? Answer should be rounded to the appropriate number of significant figures and have the correct unit.

Answer 1

The balanced chemical equation for the combustion of methane gas (CH4) is:

CH4 + 2O2 → CO2 + 2H2O

From the equation, we can see that for every mole of methane gas that reacts, 2 moles of water are produced.

The molar mass of water (H2O) is 18.01528 g/mol.

To determine the number of moles of water produced by 330 g of water, we can use the following equation:

moles = mass/molar mass

moles of water = 330 g / 18.01528 g/mol = 18.3276 mol

Since the ratio of methane to water in the balanced equation is 1:2, we know that for every mole of water produced, half a mole of methane is consumed.

moles of methane = 0.5 x moles of water = 0.5 x 18.3276 mol = 9.1638 mol

Therefore, 9.16 moles of methane gas will combust to produce 330 grams of water.