Question

c3H8 + 5O2 = 3CO2 + 4H2O When 44.0 grams of propane (C3H8) under goes complete combustion, how many grams of water will be produced?

Answer 1

Answer: 72 grams of water will be produced.

Explanation:

To calculate the number of moles, we use the formula:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$    ....(1)

Mass of propane = 44 grams

Molar mass of propane = 44 grams

Putting values in above equation, we get:

$\text{moles of propane}=\frac{44g}{44g/mol}=1mole$

For the reaction of combustion reaction of propane, the equation follows:

$C_3H_8+5O_2\rightarrow 3CO_2+4H_2O$

By Stoichiometry of the reaction,

1 mole of propane produces 4 moles of water.

So, 1 mole of propane will produce = $\frac{1}{1}\times 4=4moles$ of water.

Now, to calculate the amount of water, we use equation 1, we get:

Molar mass of water = 18 g/mol

$4mol=\frac{\text{Mass of water}}{18g/mol}$

Mass of water produced = 72 grams

Hence,  72 grams of water will be produced.