Question

Phosphorus reacts with oxygen to form diphosphorus pentoxide, P2O5 . 4P(s)+5O2(g)⟶2P2O5(s) How many grams of P2O5 are formed when 2.45 g of phosphorus reacts with excess oxygen? Show the unit analysis used for the calculation by placing the correct components into the unit-factor slots

Answer 1

5.619 grams of diphosphorus pentoxide are formed when 2.45 g of phosphorus reacts with excess oxygen.

Step-by-step explanation:

`4P(s)+5O-2(g)\rightarrow 2P_2O_5(s)`

Mass of phosphorus = 2.45 g

Moles of phosphorous =
`(2.45 g)/(31 g/mol)=0.7903 mol`

According to reaction 4 moles of phosphorus gives 2 moles of diphosphorus pentoxide.

Then 0.7903 moles of phosphorus will give:

`(2)/(4)* 0.7903 mol=0.03957 mol` of diphosphorus pentoxide

Mass of 0.03957 moles of diphosphorus pentoxide :

`0.03957 mol* 142 = 5.619 g`

5.619 grams of diphosphorus pentoxide are formed when 2.45 g of phosphorus reacts with excess oxygen.

Answer 2

The answer to your question is:

Step-by-step explanation:

2.45 g of Phosphorus

MW P = 31 g

MW P2O5 = 2(31) + 5(16) = 142 g

From the balance reaction

4 P ⇒ 2 P2O5

Then 4(31) g P ⇒ 2 (142) g P2O5

124g of P ⇒ 284 g of P2O5 Rule of three

2.45g P ⇒ x

x = 2.45 x 284/124 = 695.8/124 = 5.61 g of P2O5