Question

given that the molar mass of H2O is 18.02g/mol, how many liters of propane are required at STP to produce 75 g of H2O from this reaction?

Answer 1

C₃H₈ + 5O₂ = 3 CO₂ + 4 H₂O

44.1 g ----------- 4* 18.02 g

? g --------------- 75 g

mass of C₃H₈ = 75 * 44.1 / 4 * 18.02

mass of C₃H₈ = 3307.5 / 72.08

= 45.886 g of propane

number of moles:

45.886 / 44.1 = 1.0404 moles

1 mole --------- 22.4 L ( at STP )
1.0404 moles ----- ?

v = 1.0404 * 22.4 / 1

v = 23.304 L

hope this helps!