First, calculate the molar mass of each of the compounds. This is how much, in grams, one mole of each compound weighs. You calculate this by finding the atomic mass of each element found in a compound and multiplying that by the number of atoms present; then, add the totals for all of the elements in the molecule and bam, compound’s molar mass.

For example, for P4O10 the molar mass is about 320 g/mol. (atomic mass of P = 40 grams x 4 = 160; atomic mass of H = 16 g x 10 = 160; 160 + 160 = 320 g/mol.) PH3 is 43 g/mol, O2 is 32 g/mol, H2O is 18 g/mol. Moving on.

I am going to assume for part a) that there is an excess of O2 and that it isn’t the limiting reagent; if it isn’t, there isn’t enough info given to do the problem.

Using that assumption, part a) is about mole ratios. Note that in the original equation, PH3 is preceded by a 4; P4O10 has no number so we can assume it’s a 1; therefore, PH3 and P4O10 are in a 4:1 mole ratio. We know how many moles of PH3 we have: 12.43. Divide this by 4 to find the number of moles of P4O10 that will be produced: 3.1 (roughly; calculate this properly.) THEN, multiply that number (3.1) by the molar mass of P4O10 (320 g/mol) to find your answer: 997 g.

For part b), it’s a bit more complicated. The mole ratio is 4:6 for PH3:H2O which makes things a little harder. I always have to cross multiply fractions to figure these out…which I will try to render here:

A 4 = x C

— —

B 6 0.739 D

(edit: the formatting above failed when I posted, but I thought it might still help so I left it in. A/B is 4/6. C/D is x/0.739.)

Multiply A by D, B by C and solve for x. (Because A/B is the same ratio as C/D, this works.) It comes to about 0.203. This is the number of moles of PH3. Multiply by molar mass…(43 g/mol) and you get 8.73 g, more or less. There’s the answer for part b.

c) and d) are more of the same, but they use O2 as the limiting reagent, methinks. I think you should be able to figure them out now.

Basically, stoichiometry relies primarily on mole ratios and gram/mole conversion (which is why knowing molar masses is so important!) If you figure out those two concepts you should be pretty much set.

**********to respond to your new comments************

This is where I’d be really cocky and say that your book is wrong. My high school textbooks were absolutely riddled with errors, and if you’ve tried the problem a whole bunch of times, using different methods and you STILL can’t get an answer, the chance is solid that the book is wrong (if your books are anything like mine were.)

I can’t come up with your book’s answer either, by the way…..and I keep coming up with the value I gave above (8.73 g, roughly)….so either I’ve forgotten how to do stoichiometry OR I keep making a calculation error (I’ve got a calculator now) OR your book is wrong. If you’re turning in this question as part of an assignment, just show your work; you’ll get some part marks regardless.

This is the last time I’m going to check back to this question so I hope you figure it out! Good luck.