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How many grams of air are required to complete the combustion of 120 g of phosphorous to diphosphorous pentoxide, assuming the air to be 23% oxygen by mass?

User Dmarcato
by
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1 Answer

1 vote

Answer:

Step-by-step explanation:

522 g

Step-by-step explanation:

Your starting point here will be the balanced chemical equation for this combustion reaction

4

P

(s]

+

5

O

2(g]

2

P

2

O

5(s]

Notice that you have a

4

:

5

mole ratio between phosphorus and oxygen. This means that, regardless of how many moles of phosphorus you have, the reaction will always need

5

4

time more moles of oxygen gas.

Use phosphorus' molar mass to determine how many moles you have in that

93.0-g

sample

93.0

g

1mole P

30.974

g

=

3.0025 moles P

Use the aforementioned mole ratio to determine how many moles of oxygen you would need for many moles of phosphorus to completely take part in the reaction

3.0025

moles P

5

moles O

2

4

moles P

=

3.753 moles O

2

User Chatlanin
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