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.jpg Given that the molar mass of H2O is 18.02 g/mol, how many liters of propane are required at STP to produce 75 g of H2O from this reaction? 23 L 23.3 L 93 L 93.2 L

Answer 1

Answer:

23.30 liters of propane are required at STP to produce 75 g of water from this reaction.

Step-by-step explanation:

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

Water produced = 75 g

Moles of water =$\frac{75 g}{18.02 g/mol}=4.1620 mole$

1 mole of propane gives 4 moles of water.

Then, 4.1620 moles of water will be obtained form:

$\frac{1}{4}\times 4.1620=1.0405 mole$ of propane

At STP ,1 mole of gas occupies = 22.4 L

Then, 1.0405 mole will occupy:

$1.0405\times 22.4 L=23.30 L$

23.30 liters of propane are required at STP to produce 75 g of water from this reaction.

Answer 2

23L (A) is the correct answer