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25 December, 01:56

P4O10 + 6PCl5---> 10 POCl3

How many grams of POCl3 are produced when 225.0g of P4O10 and 675.0g of PCl5 react?

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  1. 25 December, 04:01
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    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

    Answer: 997.8 g
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