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Chapter 9: Problem 79

Consider the following unbalanced chemical equation: \(\mathrm{P}+\mathrm{O}_{2} \rightarrow \mathrm{P}_{2} \mathrm{O}_{5}\) (a) How many grams of phosphorus (P) are required to react completely with \(20.0 \mathrm{~g}\) of \(\mathrm{O}_{2} ?\) (b) What is the theoretical yield in grams if you combine the amounts of reactants in part (a)?

Short Answer

Expert verified
(a) 15.5 g of phosphorus are required to react completely with 20.0 g of O₂. (b) The theoretical yield of P₂O₅ formed by combining the reactants in part (a) is 35.5 g.

Step by step solution

01

Balance the chemical equation

First, balance the equation by determining the number of atoms of each element on both sides of the equation. The balanced equation of the reaction between phosphorus and oxygen is: \[4\mathrm{P} + 5\mathrm{O}_{2} \rightarrow 2\mathrm{P}_{2}\mathrm{O}_{5}\]
02

Convert given mass to moles

Given that there are 20.0 g of O₂. Start by converting this mass to moles using the molar mass of O₂ (32.00 g/mol): Moles of O₂ = \(\frac{\text{20.0 g}}{\text{32.00 g/mol}}=\) \(0.625 \, \text{mol}\)
03

Use the stoichiometry of the balanced equation to calculate moles of phosphorus needed

Use the stoichiometric coefficients in the balanced equation to determine the moles of phosphorus needed: Moles of P = (moles of O₂) \(\times \frac{\text{4 moles of P}}{\text{5 moles of O₂}}\) Moles of P = \((0.625 \, \text{mol}) \times \frac{4}{5} = 0.5 \, \text{mol}\)
04

Convert the moles of phosphorus to grams

Now, convert moles of phosphorus into grams using its molar mass (30.97 g/mol): Grams of P = (moles of P) \(\times\) (molar mass of P) Grams of P = \((0.5 \, \text{mol}) \times \text{30.97 g/mol}\) = \(15.5 \, \text{g}\) (a) 15.5 g of phosphorus are required to react completely with 20.0 g of O₂.
05

Calculate the theoretical yield of P₂O₅

The balanced equation shows that 4 moles of phosphorus react with 5 moles of oxygen to form 2 moles of P₂O₅. Determine the moles of P₂O₅ formed: Moles of P₂O₅ = (moles of P) \(\times \frac{\text{2 moles of P₂O₅}}{\text{4 moles of P}}\) Moles of P₂O₅ = \((0.5 \, \text{mol}) \times \frac{2}{4} = 0.25 \, \text{mol}\) Now, convert moles of P₂O₅ into grams using its molar mass (141.94 g/mol): Grams of P₂O₅ = (moles of P₂O₅) \(\times\) (molar mass of P₂O₅) Grams of P₂O₅ = \((0.25 \, \text{mol}) \times \text{141.94 g/mol}\) = \(35.5 \, \text{g}\) (b) The theoretical yield of P₂O₅ formed by combining the reactants in part (a) is 35.5 g.

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