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.

Homework Answers

Answer #1

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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