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.

Step-by-step 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}=(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 =
`(1)/(1)* 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.