what weight of each substance is present after 0.4500g of p4o10 and 1.5g of pcl5 are reacted completely
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You need to find the limiting reagent. Do some stoichiometric calculations to find out how many mols of P4O10 we would need to react with 1.5 g PCl5 (call this a) and how many mols of PCl5 we would need to react with .45 g of P4O10 (call this b)
If a is less than .45 g P4O10, PCl5 is limiting and we will have extra P4O10. If b is less than 1.5 g PCl5, P4O10 is limiting and we will have extra PCl5.
Use whichever reactant turns out to be the limiting reagent to stoichiometrically calculate how much POCl3 is formed. The limiting reagent will have weight of 0. For the other reactant's weight, just take the original .45 g and subtract a if PCl5 was limiting or take 1.5 and subtract b if P4O10 was limiting.
Good luck!
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