Question

A sample of 3.62 moles of diphosphorous trioxide is
reacted with 6.31 moles of water to produce
phosphorous acid. Using the balanced equation
below, predict which is the limiting reactant and the
maximum amount in moles of phosphorous acid that
can be produced.
P2O3 + 3H₂O → 2H3PO3
A. water, 8.42 moles
B. water, 4.21 moles
C. diphosphorus trioxide, 7.22 moles
D. diphosphorus trioxide, 3.75 moles

Answer 1

$H_3PO_3$$2H_3PO_3$$2H_3PO_3$$P_2O_3$Answer:

B

Explanation:

This question is about stoichiometry. From the balanced equation $P_2O_3 + 3H_2O$⇒$2H_3PO_3$, we see that 3 moles of water is needed to react with 1 mole of $P_2O_3$.

This means that, to fully react 3.62 moles of $P_2O_3$, we would need 3*3.62 or  10.86 moles of water. However, we only have 6.31 moles, so water is the limiting reactant.

Since 3 moles of water react with 1 mole of $P_2O_3$, 6.31 moles of water can fully react with 6.31÷3 or 2.1033 moles of $P_2O_3$.

From the balanced equation, we see that every mole of $P_2O_3$ reacted gets you 2 moles of $2H_3PO_3$. Therefore, 2.1033 moles of $P_2O_3$ would give you approximately 4.21 moles of $H_3PO_3$.