Consider the combustion of 30.0 g of propane in excess oxygen. Write the balanced molecular equation for this reaction and find the maximum mass of H2O that can form

Answer 1

Hi Brittney V.,

C₃H₈ + O₂ ---> CO₂ + H₂O, from her I see I need 4H₂ to make H₈, and 3C to make C₃,

C₃H₈ + 5O₂ ---> 3CO₂ + 4H₂O, and 5O₂ balances it.

C₃H₈ is 3*(12.0107g) + 8*(1.00794g) = 44.0956 g/mol

H₂O is 2*(1.00794g) + 1*(15.9994g) = 18.0153g/mol

So how many g's H2O = 30.0g C₃H₈ ,

30.0 g C₃H₈ *(1 mol C₃H₈)*(4 mol H₂O)*(18.0153 g H₂O) = 49.026 g H₂O or 49.0 g H₂O (sig. fig.)

(44.0956 g C₃H₈) *(1 mol C₃H₈)*(1 mol H₂O)

Check my work. It's been a while since I've done chemistry.

I hope this helps, Joe.

Answer 2

Balanced equation: C₃H₈ + 5O₂ ==> 3CO₂ + 4H₂O

moles propane present = 30.0 g x 1 mole propane/44.1 g = 0.680 moles

max. mass H₂O produced = 0.680 moles C₃H₈ x 4 moles H₂O/mole C₃H₈ x 18.0 g/mole H₂O = 49.0 g H₂O