Question

1C3H8 + 5O2 = 3CO2 + 4H2O

C3H8 = 100g
5O2 = 60g

Which reactant above is the limiting reactant?

How many grams of CO2 are produced?

How many molecules of CO2 are produced?

Answer 1

C3H8 + 5O2 = 3CO2 + 4H2O a) moles C3H8 present = 14.8 g x 1 mol/44 g = 0.336 moles moles O2 present = 3.44 g x 1 mole/32 g = 0.1075 From balanced equation

Step-by-step explanation:

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