Question

# Consider a situation in which 235 g of P4 are exposed to 272 g of O2....

Consider a situation in which 235 g of P4 are exposed to 272 g of O2.

In Part A, you found the number of moles of product (3.80 mol P2O5 ) formed from the given amount of phosphorus and excess oxygen. In Part B, you found the number of moles of product (3.40 mol P2O5 ) formed from the given amount of oxygen and excess phosphorus.

Now, determine the number of moles of P2O5 is produced from the given amounts of phosphorus and oxygen.

Reaction :-

P4 + 5O2 -----> 2P2O5

Mass of P4 = 235 g, Molar mass of P4 = 123.90 g/mol

Number of moles of P4 = 235 g / 123.90 g/mol = 1.897 mol

Mass of O2 = 272 g, Molar mass of O2 = 32 g/mol

Number of moles of O2 = 272 g / 32 g/mol = 8.5 mol

From reaction,

5.0 mol of O2 reacts with 1.0 mol of P4 produces 2.0 mol of P2O5 so 8.5 mol of O2 will react with 1.0 * 8.5 mol / 5.0 = 1.7 mol of P4 will produce 2*1.7 mol = 3.40 mol of P2O5.

Since P4 have not sufficient moles to react compltely hence it will react with whole moles of O2.

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