Question

From the reaction: CH4 (g) + O2 (g) ► CO2 (g) + 2H2O (l).

A) What mass of O2 will be needed to burn 36.1 g of CH4?
B) How many moles of water are produced from 19.2 g of O2?

Answer 1

To burn 36.1 g of CH4, 144 g of O2 are needed. From 19.2 g of O2, 1.20 mol of H2O are produced.

Step-by-step explanation:

The question presents two separate problems related to the combustion of methane (CH4).

Part A:

To find the mass of O2 needed to burn 36.1 g of CH4, we need to use mole-mass calculations. The molar mass of CH4 is approximately 16.04 g/mol. From the stoichiometry of the balanced chemical equation, the molar ratio of CH4 to O2 is 1:2. If we have 36.1 g of CH4, this is equivalent to 36.1 g / 16.04 g/mol = 2.25 mol of CH4. Therefore, we need 2.25 mol * 2 = 4.50 mol of O2. The molar mass of O2 is approximately 32.00 g/mol, so the mass of O2 required is 4.50 mol * 32.00 g/mol = 144 g.

Part B:

To determine the number of moles of water produced from 19.2 g of O2, again we use stoichiometry. Since the molar mass of O2 is 32.00 g/mol, 19.2 g corresponds to 19.2 g / 32.00 g/mol = 0.60 mol of O2. According to the balanced equation, 1 mol of O2 produces 2 mol of H2O. Therefore, 0.60 mol of O2 will result in 0.60 mol * 2 = 1.20 mol of H2O.