P4o10 + > 10 pocl3
how many grams of pocl3 are produced when 225.0g of p4o10 and 675.0g of pcl5 react?

P4O10 + 6PCl5 --> 10 POCl3

That equation is balances (same number of atoms of each element on the left than on the right).

1) Writhe the stechiometric proportions:

     1 mol P4O10 : 6mol PCl5 : 10 mol POCl3

2) Calculate the molar masses of each compound (using the aomic masses of the elements)

     molar mass of P4O10 = 4*31g/mol + 10*16g/mol = 284 g/mol

     molar mass of PCl5 = 31g/mol * 5*35.5 g/mol =  173 g/mol

     molar mass of POCl3 = 31 g/mol + 16g/mol + 3*35.5g/mol =  153.5 g/mol

3) Pass the information in grams to mol by dividing grams / molar mass

     225.0 g P4O10 / 284 g/mol = 0.792 mol P4O10

     675.0 g PCl5 / 173 g/mol = 3.902 mol PCl5

4) Find the limitant reagent

     ratio 3.902 mol PCl5 / 0.792 mol P4O10 = 4.93

      4.93 < 6/1 which is the theoretical ratio, then there is less PCl5 and it will be consumed before ending the P4O10.

5) Use the # of moles of the limitan reagent to find the # of moles of the product, using the theoretical proportions

[10 mol POCl3 / 6mol PCl5 ] *  3.902 mol PCl5 =6.50 mol POCl3

6) Pass the number of moles of POCl3 to grams, using its molar mass

6.50 mol POCl3 * 153.5 g/mol = 997.8 grams 

997.8 g

Could you make it more clear?

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