Question

Using the equation below, calculate how many grams of water can be produced by complete combustion of 5.51g of butane in excess oxygen?
2C4H10+13O2→8CO2+10H2O

Answer 1

Answer:

answer in attachment

Answer 2

Given - Amount of butane - 5.51 gram

Find - Amount of water produced from given amount of butane.

Solution - From the chemical reaction in question, 2 moles of butane produces 10 moles of water by complete combustion in presence of oxygen.

Number of moles of butane - 5.51/58.12

Number of moles of butane - 0.09

0.09 moles of butane will form water - 0.47 moles

Weight of water produced - 0.47*18

Weight of water produced - 8.46 gram

Hence, 5.51 gram of butane will form 8.46 gram of water.