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?

Explanation:

522 g

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

weight of any object is given by

w = m*g

where m is mass of the object and g is the acceleration due to gravity.

for earth the value of g is 9.8 m/s^2 so

w=117.6 n m=12 g=?

117.6=12g => g=9.8 m/s^2

which is the force of gravity on earth.