Question

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?

Answer 1

we need a balanced equation:

C₃H₈ + 5O₂ ---> 3CO₂ + 4H₂O

let's convert the grams of water to moles using the molar mass. then mole-mole ratio to convert to moles of propane. finally, convert moles to Liters using the STP molar volume (1 mol= 22.4 L)


`75 g H_2O ( (1 mol H_2O)/(18.02 grams) ) ( (1 mol C_3H_8)/(4 mol H_2O) ) ( (22.4+ L)/(1 mol) )= 23.3 L`