Chemistry - Stoichiometry

Question

C3H8 + 5 O2 --> 4 H2O + 3 CO2

In this equation, how many grams of O2 is required to produce 720 grams of H2O?

I know that first we need to calculate the moles for H2O and from it we can get the moles for O2 and then the weight but my problem is when I try to calculate the moles of O2. I'm unsure how to do it.

Answer 1

Your outlined approach is correct. Let's see if we can explain how to accomplish this.

C₃H₈ + 5O₂ ==> 4H₂O + 3CO₂ ... balanced equation

So as you state, we will first calculate moles of H₂O:

...... 720 g H₂O x 1 mol H₂O/18 g = 40.0 moles H₂O

And again, as you state, we next find moles of O₂ needed to produce this amount of water:

...... 40.0 moles H₂O x 5 moles O₂/4 moles H₂O = 50 moles of O₂ required.

This is easily seen from looking at the balanced equation and the mole ratio (stoichiometry) between O₂ and H₂O. It takes 5 moles O₂ to produce 4 moles of H₂O. This is where the 40 mole H₂O x 5 mole O₂ / 4 moles H₂O comes from.

Finally, we convert 50 moles of O2 to grams:

...... 50 moles O₂ x 32 g/mole = 1600 grams = 1.60x10³ grams (to 3 sig. figs.)