CountryGirl59547 CountryGirl59547
  • 13-05-2016
  • Chemistry
contestada

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?

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Willchemistry
Willchemistry Willchemistry
  • 13-05-2016

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!
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