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.
[tex]\text{Moles of water}=\frac{\text{Mass of water}}{\text{Molar mass of water}}=\frac{75g}{18g/mole}=4.16moles[/tex]
Now we have to calculate the moles of propane.
The balanced chemical reaction will be,
[tex]C_3H_8+5O_2\rightarrow 4H_2O+3CO_2[/tex]
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 [tex]\frac{4.16}{4}=1.04[/tex] 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 [tex]22.4L\times 1.04=23.29L[/tex] volume of propane gas
Therefore, the volume of propane required are, 23.29 L
