Phosphoric acid (H3PO4, molar mass 98.0) can be prepared from P4 and O2 by the following series of reactions:
P4 + 5O2 ? P4O10
P4O10 + 6H2O ? 4H3PO4
16.5 grams of H3PO4 were formed by reacting 0.0500 mole of P4 with sufficient O2 and H2O. Calculate the percent yield.
The Correct Answer and Explanation is :
To calculate the percent yield of phosphoric acid (H₃PO₄), we need to follow a few steps:
Step 1: Moles of Phosphoric Acid (H₃PO₄) Expected
From the balanced reaction:
[ P_4 + 5O_2 \to P_4O_{10} ]
[ P_4O_{10} + 6H_2O \to 4H_3PO_4 ]
We are given that 0.0500 mole of P₄ reacts to form H₃PO₄. The second reaction shows that 1 mole of P₄O₁₀ (formed from 1 mole of P₄) produces 4 moles of H₃PO₄. Therefore, we can calculate the theoretical amount of H₃PO₄:
[
\text{Moles of H₃PO₄} = 0.0500 \, \text{mol of P₄} \times \frac{4 \, \text{mol of H₃PO₄}}{1 \, \text{mol of P₄}} = 0.200 \, \text{mol of H₃PO₄}
]
Step 2: Theoretical Mass of H₃PO₄
Now, we can calculate the theoretical mass of H₃PO₄ produced:
[
\text{Mass of H₃PO₄} = 0.200 \, \text{mol of H₃PO₄} \times 98.0 \, \text{g/mol} = 19.6 \, \text{g}
]
Step 3: Percent Yield
The percent yield is calculated using the formula:
[
\text{Percent Yield} = \frac{\text{Actual Yield}}{\text{Theoretical Yield}} \times 100
]
We are given that the actual yield is 16.5 grams. So:
[
\text{Percent Yield} = \frac{16.5 \, \text{g}}{19.6 \, \text{g}} \times 100 = 84.2\%
]
Explanation
In this problem, the theoretical yield of phosphoric acid (H₃PO₄) is determined by stoichiometry, where we use the number of moles of the reactant (P₄) to predict how much product (H₃PO₄) should be formed under ideal conditions. Since we know the actual yield of H₃PO₄ is 16.5 grams, the percent yield measures the efficiency of the reaction by comparing the actual yield to the theoretical yield. In this case, the reaction achieved 84.2% of the expected product, which indicates a relatively efficient synthesis of phosphoric acid.