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A sample of 82.3 g of tetraphosphorus decoxide (P₄O₁₀) reacts with 66.9 g of water to produce phosphoric acid (H₃PO₄) according to the following balanced equation:

\[ \text{P}_4\text{O}_{10} + 6\text{H}_2\text{O} \rightarrow 4\text{H}_3\text{PO}_4 \]

Determine the limiting reactant for the reaction.

Answer :

To determine the limiting reactant for the given reaction, calculate the moles of each reactant from their masses. The chemical equation shows that tetraphosphorus decoxide is the limiting reactant because there is more than enough water to react with it.

To determine the limiting reactant for the reaction between tetraphosphorus decoxide (P4O10) and water (H2O) to produce phosphoric acid (H3PO4), one must first calculate the moles of each reactant based on their given masses and molar masses. The balanced chemical equation is P4O10 + 6H2O ⇒ 4H3PO4.

The molar mass of P4O10 is approximately 283.9 g/mol (calculated as 4 P atoms times the atomic weight of phosphorus plus 10 O atoms times the atomic weight of oxygen), so 82.3 g of P4O10 corresponds to

82.3 g / 283.9 g/mol = 0.290 moles.

The molar mass of H2O is 18 g/mol, so 66.9 g of H2O corresponds to

66.9 g / 18 g/mol = 3.72 moles.

According to the stoichiometry of the balanced equation, 1 mole of P4O10 reacts with 6 moles of H2O.

Therefore, to completely react with 0.290 moles of P4O10, we would need

0.290 moles x 6 = 1.74 moles of H2O.

Since we have 3.72 moles of H2O, which is more than the required 1.74 moles, H2O is in excess, and P4O10 is the limiting reactant.

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