Question

The combustion of liquid octane C8H18, in the presence of gaseous oxygen yields carbon dioxide and liquid water. How many grams of water are produced by the reaction of 18.0 g of octane?

Answer 1

To determine the number of grams of water produced by the combustion of 18.0 g of octane, we need to balance the chemical equation, calculate the number of moles of octane, use the mole ratio to calculate the number of moles of water, and then convert moles of water to grams.

Step-by-step explanation:

To determine the number of grams of water produced by the combustion of 18.0 g of octane, we first need to balance the chemical equation:

2 C8H18 + 25 O2 → 16 CO2 + 18 H2O

From the balanced equation, we can see that for every 2 moles of octane, 18 moles of water are produced. Now we can use the molar mass of octane (114 g/mol) to calculate the number of moles of octane in 18.0 g:

moles of octane = 18.0 g / 114 g/mol = 0.157 moles

Finally, we can use the mole ratio between octane and water to calculate the number of moles of water produced:

moles of water = 0.157 moles * (18 moles H2O / 2 moles C8H18)

moles of water = 1.41 moles

To convert moles of water to grams, we can use the molar mass of water (18.02 g/mol):

grams of water = 1.41 moles * 18.02 g/mol = 25.4 g

Therefore, 25.4 grams of water are produced by the combustion of 18.0 g of octane.