# Video: Calculating the Mass of Water Produced by the Complete Combustion of a Given Amount of Butane, Given the Balanced Chemical Reaction Equation

Butane burns in air according to the following balanced equation: C₄H₁₀ (g) + (13/2)O₂ (g) ⟶ 4CO₂ (g) + 5H₂O (g). What mass of water, in grams, will be produced from the complete combustion of 2.0 moles of butane? [A] 18 grams [B] 22 grams [C] 72 grams [D] 90 grams [E] 180 grams.

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### Video Transcript

Butane burns in air according to the following balanced equation: C4H10 gas plus 13 over two O2 gas giving four CO2 gas plus 5H2O gas. What mass of water, in grams, will be produced from the complete combustion of 2.0 moles of butane? (A) 18 grams, (B) 22 grams, (C) 72 grams, (D) 90 grams, or (E) 180 grams.

Butane, C4H10, is a hydrocarbon gas at room temperature. Combustion is a type of reaction with oxygen gas and usually produces energy in the form of heat and light. So, we could write plus energy at the end of the reaction equation. And we call this an exothermic reaction because a net release of energy occurs. Butane, usually mixed with propane, is a fuel used in gas stoves and gas canisters for cooking. It is also a component of the gas mixture used to fuel Bunsen burners.

When butane burns in air, we can assume that oxygen gas is present in excess quantities, and thus butane is the limiting reagent. This is important to know because it is the number of moles of limiting reagent which determines the number of moles of product that can form. For every one mole of butane gas that combusts, four moles of CO2 gas and five moles of gaseous water can be produced. This is assuming the complete combustion of butane, which means all the butane reacts.

The question asks for the mass of water that can be produced. So, we will only consider butane and water. We know the reaction ratio or mole ratio of butane as to water is one as to five. We can take the number of moles of butane given to us, 2.0, and calculate the number of moles of water that can be produced. The value of 𝑥 will be five times the number of moles of butane. So, five times the number of moles of butane is equal to five times 2.0 moles, which gives us 10.0 moles of water which can be produced.

Next, we need to calculate the mass of water from its number of moles. We take the number of moles of water, which we have just calculated, 10.0 moles. And using the equation number of moles as mass of a molar mass which we can rearrange to get mass is equal to number of moles times molar mass. We can plug in the number of moles of water, which we now have. We need to now calculate the molar mass of water and put this value into the equation.

The molar mass of water is calculated as follows. Take the molar mass of hydrogen, which is one grams per mole, and multiply it by two because there are two hydrogens. And add the molar mass of the one oxygen, which is 16 grams per mole. And we get a molar mass for water of 18 grams for every mole of water. Let’s put that into our equation, and we get an answer of 180 grams of water.

Finally, the mass of water in grams that can be produced from the complete combustion of 2.0 moles of butane is 180 grams.