Question

Propane (C3H8), a fuel that is used in camp stoves, produces carbon dioxide (CO2) and water vapor (H2O) on combustion as follows. Mc032-1. Given that the molar mass of H2O is 18.02 g/mol, how many liters of propane are required at stp to produce 75g of H2O from this reaction

Answer 1

Answer : The volume of propane required are, 23.29 L

Solution : Given,

Mass of water = 75 g

Molar mass of water = 18 g/mole

First we have to calculate the moles of water.

$\text{Moles of water}=\frac{\text{Mass of water}}{\text{Molar mass of water}}=\frac{75g}{18g/mole}=4.16moles$

Now we have to calculate the moles of propane.

The balanced chemical reaction will be,

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

From the reaction, we conclude that

As, 4 moles of water produces from 1 mole of propane

So, 4.16 moles of water produces from $\frac{4.16}{4}=1.04$ mole of propane

Now we have to calculate the volume of propane.

As, 1 mole of propane contains 22.4 L volume of propane gas

So, 1.04 mole of propane contains $22.4L\times 1.04=23.29L$ volume of propane gas

Therefore, the volume of propane required are, 23.29 L

Answer 2

23L (A) is the correct answer :L