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

Answer:

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

Explanation:

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

Mass of  phosphorus = 2.45 g

Moles of phosphorous = $\frac{2.45 g}{31 g/mol}=0.7903 mol$

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

Then 0.7903  moles of phosphorus will give:

$\frac{2}{4}\times 0.7903 mol=0.03957 mol$ of diphosphorus pentoxide

Mass of 0.03957 moles of  diphosphorus pentoxide :

$0.03957 mol\times 142 = 5.619 g$

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

Answer 2

Answer:

The answer to your question is:

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