Question

What is the maximum amount of P2O5 that can theoretically be made from 161 g of P4 and excess oxygen? I got 2.60 mol

What is the maximum amount of P2O5 that can theoretically be made from 176 g of O2 and excess phosphorus? 2.20 mol

Last is What is the percent yield if the actual yield from this reaction is 256 g ?

Answer #1

The balanced equation is P_{4} + 5O_{2}
2P_{2}O_{5}

Molar mass of P_{4} = 4xAt.mass of P = 4x31 = 124
g/mol

Molar mass of P_{2}O_{5} = (2xAt.mass of P ) + (
5xAt.mass of O )

= (2x31) + (5x16)

= 142 g/mol

According to the balanced equation ,

1 mole of P_{4} produces 2 moles of
P_{2}O_{5}

OR

1x124 g of P_{4} produces 2x142 g of
P_{2}O_{5}

161 g of P_{4} produces M g of
P_{2}O_{5}

M = ( 161 x 2x142) / (1x124)

= 368.7 g of P_{2}O_{5}

Therefore the maximum amount of
P_{2}O_{5} formed is 368.7 g

percent yield = ( actual yield / theoretical yield) x 100

= ( 256 / 368.7 ) x100

= 69.4 %

-----------------------------------------------------------------------------------------------

According to the balanced equation ,

1 mole of P_{4} produces 2 moles of
P_{2}O_{5}

OR

1x124 g of P_{4} produces 2x142 g of
P_{2}O_{5}

176 g of P_{4} produces M g of
P_{2}O_{5}

M = ( 176 x 2x142) / (1x124)

= 403.1 g of P_{2}O_{5}

Therefore the maximum amount of
P_{2}O_{5} formed is 403.1 g

percent yield = ( actual yield / theoretical yield) x 100

= ( 256 / 403.1 ) x100

= 63.5 %

P4+5O2→2P2O5
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