how many grams of water are produced when 34.6 grams of water are produced when 34.6 grams of butane react with excess oxygen gas?

Answer 1

To determine the mass of water produced when 34.6 grams of butane (C4H10) reacts with excess oxygen gas (O2), we need to use stoichiometry and the balanced equation for the combustion of butane.

The balanced equation for the combustion of butane is:

2 C4H10 + 13 O2 → 8 CO2 + 10 H2O

From the balanced equation, we can see that the stoichiometric ratio between butane and water is 2:10, which simplifies to 1:5. This means that for every 1 mole of butane reacted, 5 moles of water are produced.

First, let's calculate the number of moles of butane using its molar mass:

Molar mass of C4H10 = (4 * 12.01 g/mol) + (10 * 1.01 g/mol) = 58.12 g/mol

Moles of butane = 34.6 g / 58.12 g/mol ≈ 0.595 moles

Since the stoichiometric ratio between butane and water is 1:5, the number of moles of water produced is 5 times the number of moles of butane:

Moles of water produced = 5 * 0.595 moles = 2.975 moles

Now, we can calculate the mass of water produced using the molar mass of water:

Molar mass of H2O = (2 * 1.01 g/mol) + 16.00 g/mol = 18.02 g/mol

Mass of water produced = 2.975 moles * 18.02 g/mol ≈ 53.61 grams

Therefore, approximately 53.61 grams of water are produced when 34.6 grams of butane react with excess oxygen gas.