Hello, Nabil,
Make a table with what you know. Calculate the number of moles of O2 and H2O from the masses given. We'll assume there is an excess of C3H8, so all of the O2 is consumed. We know from the balanced equation that we should get 4 moles of water for every 5 moles of O2, or a molar ratio of 0.80.
From the table, we see we started with 0.388 moles of O2. Multiply that by the molar ratio (o.80) to calculate how many moles of water we would expect, if everything went hunky dory (metric for 100% yield). We would expect 0.31 moles, or 5.58 grams. Instead, we only got 3.25 grams, thanks to your lab partner, I assume.
Percent theorectical yield is thus 3.25g/5.58g, or 0.58. That's 58, without the percent sign, as requested.
58% for the rest of us.
Bob
