Calculate the mass of water produced when 2.64 g of butane reacts with excess oxygen.

The reaction of butane with sufficient/excess oxygen is as follows:

2C₄H₁₀ + 13O₂ ⇒ 8CO₂ + 10H₂O

To work out how much water is produced, we first need to convert 2.64g of butane to moles by dividing it by butane's molecular mass, which is 58g/mol. From this we can see that we have roughly 0.045mol of butane. Since 0.045 is 0.022 of 2, we multiply this to get 0.22mol of water produced. The last thing to do is to multiply this by 18g (the molecular mass of water) to get 3.96g of water produced.