Question

P4O10 + 6PCl5---> 10 POCl3
How many grams of POCl3 are produced when 225.0g of P4O10 and 675.0g of PCl5 react? How would I find this using dimensional analysis!

Answer 1

no. of mole of P4O10 used= 225.0/(31*4+16*10) = 0.79225 ; no. of mole of PCl5 used=675/(31+35.5*5) = 3.23741mole
no of mole of PCl5 required to react completely with 225.0g of P4O10 = 0.79225 *6 = 4.7535 mole. therefore P4O10 is in excess.
no. of mole of POCl3 produced = 10* (3.23741/6)=5.3957 mole. therefore mass of POCl3 required = 5.3957*(31+16+35.5*3) = 828g