Determine the theoretical yield of P2O5, when 3.07 g of P reacts with 6.09 g of oxygen in the following chemical equation 4 P+5O_2→2P_2 O_5

7.03g

Explanation:

4P + 5O2 → 2P2O5

Let us convert the mass given to mole. This can be achieved by doing the following:

Molar Mass of O2 = 16x2 = 32g/mol

Mass of O2 = 6.09g

Number of mole = Mass /Molar Mass

Number of mole of O2 = 6.09/32 = 0.19mol

Molar Mass of P = 31g/mol

Mass of P = 3.07g

Number of mole of P = 3.07/31 = 0.099mol

Let us determine the limiting reactant and the excess reactant

From the equation,

4moles of P required 5moles of O2.

Therefore, 0.099mol of P will require = (0.099 x 5)/4 = 0.12mol

From the above illustration, we see clearly that not all the O2 reacted as the number of mole of O2 obtained from the question is 0.19mol. This means that O2 is the excess reactant and P is the limiting reactant.

Note: the limiting reactant is always used to obtain the yield of any reaction.

Now we can obtain the theoretical yield of P2O5 as follows:

4P + 5O2 → 2P2O5

Molar Mass of P = 31g/mol

Mass of P from the equation = 4x31 = 124g

Molar Mass of P2O5 = (31x2) + (16x5) = 62 + 80 = 142g/mol

Mass of P2O5 from the equation = 2 x 142 = 284g

From the equation,

124g of P produced 284g of P2O5.

Therefore, 3.07g of P will produce = (3.07x284)/124 = 7.03g of P2O5.

Therefore, the theoretical yield of P2O5 is 7.03g

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