Question

Phosphorus burns in air to produce a phosphorus oxide in the following reaction- $4P + 5{O_2} \to {P_4}{O_{10}}$. What mass of phosphorus will be needed to produce 3.25 mol of ${P_4}{O_{10}}$? If $0.489$ mol of phosphorus burns, what mass of oxygen is used? What mass of ${P_4}{O_{10}}$ is produced?

Answer 1

The numerical relationship between reactants and products are known as stoichiometry. A common type of stoichiometric relationship is the mole ratio which relates the number of moles of any two substances in a chemical reaction. For the given question, first find the mole ratio for each condition and then the moles of the unknowns can be converted into mass with the help of molar mass.

Complete answer:
When phosphorus burns in excess of air, the combustion reaction takes place as follows:
$4P + 5{O_2} \to {P_4}{O_{10}}$
Now, interpreting mole ratio on the basis of stoichiometric coefficients of the given chemical reaction, for each given part:
a) mass of phosphorus will be needed to produce 3.25 mol of ${P_4}{O_{10}}$-
Since, 1 mole of ${P_4}{O_{10}}$ is produced on combustion of $\Rightarrow$ 4 moles of phosphorus
Therefore, 3.25 moles of ${P_4}{O_{10}}$ will be produced on combustion of $\Rightarrow 4 \times 3.25 = 13$ moles of phosphorus.
Molar mass of phosphorus $= 31\;gmo{l^{ - 1}}$
Hence, mass of phosphorus consumed can be calculated with the help of relation of number of moles $n = \dfrac{{{\text{given mass}}}}{{{\text{molar mass}}}}$. Substituting values:
Mass of phosphorus consumed in the given reaction condition$= 31 \times 13 \Rightarrow 403g$
b) mass of oxygen used on combustion of 0.489 mol of phosphorus-
Since, 4 moles of phosphorus burns in the presence of $\Rightarrow$ 5 moles of oxygen
Therefore, 0.489 moles of phosphorus will burn in presence of $\Rightarrow \dfrac{5}{4} \times 0.489 = 0.61$ moles of oxygen.
Molar mass of oxygen $= 32\;gmo{l^{ - 1}}$
Hence, mass of oxygen used can be calculated with the help of relation of number of moles $n = \dfrac{{{\text{given mass}}}}{{{\text{molar mass}}}}$. Substituting values:
Mass of oxygen used under given reaction conditions $= 32 \times 0.61 \Rightarrow 19.56g$
c) mass of ${P_4}{O_{10}}$ produced-
To find the mass of ${P_4}{O_{10}}$, we first need to check the limiting reagent in the reaction. As for 0.489 moles of phosphorus, 0.61 moles of oxygen is used. So, the limiting reagent for the given reaction is phosphorus.
Since, 4 moles of phosphorus burns to form $\Rightarrow$ 1 mole of ${P_4}{O_{10}}$
Therefore, 0.489 moles of phosphorus will burn in presence of $\Rightarrow \dfrac{1}{4} \times 0.489 = 0.15$ moles of ${P_4}{O_{10}}$
Molar mass of ${P_4}{O_{10}} = 284gmo{l^{ - 1}}$
Hence, substituting values in the relation $n = \dfrac{{{\text{given mass}}}}{{{\text{molar mass}}}}$:
Mass of ${P_4}{O_{10}}$ produced $= 0.15 \times 284 \Rightarrow 42.6g$

Note:
Remember that the reagent which is completely exhausted in a chemical reaction is known as limiting reagent. The quantity of limiting reagent limits the amount of product formed and hence decides when the chemical reaction will stop. In combustion reactions, the limiting reagent is generally the reactant other than oxygen because in most of the combustion reactions, oxygen is taken in excess.