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Chapter 22: Problem 56

Dinitrogen pentoxide is a product of the reaction between \(\mathrm{P}_{4} \mathrm{O}_{10}\) and \(\mathrm{HNO}_{3}\). Write a balanced equation for this reaction. Calculate the theoretical yield of \(\mathrm{N}_{2} \mathrm{O}_{5}\) if \(79.4 \mathrm{~g}\) of \(\mathrm{P}_{4} \mathrm{O}_{10}\) are reacted with an excess of \(\mathrm{HNO}_{3}\). (Hint: One of the products is \(\mathrm{HPO}_{3}\).)

Short Answer

Expert verified
The balanced equation for the reaction is \(\mathrm{P}_{4} \mathrm{O}_{10} + 12\mathrm{HNO}_{3} \rightarrow 2\mathrm{N}_{2} \mathrm{O}_{5} + 4\mathrm{HPO}_{3}\) and the theoretical yield of \(\mathrm{N}_{2} \mathrm{O}_{5}\) is \(60.3 \mathrm{g}\)

Step by step solution

01

Write Unbalanced equation

Start by writing the unbalanced chemical equation. This would be \(\mathrm{P}_{4} \mathrm{O}_{10} + \mathrm{HNO}_{3} \rightarrow \mathrm{N}_{2} \mathrm{O}_{5} + \mathrm{HPO}_{3}\)
02

Balance the equation

Balance the chemical equation. When balanced, it gives: \(\mathrm{P}_{4} \mathrm{O}_{10} + 12\mathrm{HNO}_{3} \rightarrow 2\mathrm{N}_{2} \mathrm{O}_{5} + 4\mathrm{HPO}_{3}\)
03

Calculate molar mass of P4O10 and N2O5

Calculate the molar mass of \(\mathrm{P}_{4} \mathrm{O}_{10}\) and \(\mathrm{N}_{2} \mathrm{O}_{5}\). The molar mass of \(\mathrm{P}_{4} \mathrm{O}_{10} = 283.9 \mathrm{~g/mol}\) and of \(\mathrm{N}_{2} \mathrm{O}_{5} = 108.01 \mathrm{~g/mol}\)
04

Use stoichiometry to calculate theoretical yield

According to the balanced equation, 1 mol of \(\mathrm{P}_{4} \mathrm{O}_{10}\) forms 2 mol \(\mathrm{N}_{2} \mathrm{O}_{5}\). Convert given mass of \(\mathrm{P}_{4} \mathrm{O}_{10}\) to moles by dividing by its molar mass, then multiply by 2 to find moles of \(\mathrm{N}_{2} \mathrm{O}_{5}\), and finally convert back to grams by multiplying with its molar mass which gives the theoretical yield. \((79.4 \mathrm{g} \div 283.9 \mathrm{g/mol}) \times 2 \times 108.01 \mathrm{g/mol} = 60.3 \mathrm{g}\)
05

Conclusion

The balanced chemical equation for the above reaction is \(\mathrm{P}_{4} \mathrm{O}_{10} + 12\mathrm{HNO}_{3} \rightarrow 2\mathrm{N}_{2} \mathrm{O}_{5} + 4\mathrm{HPO}_{3}\) and the theoretical yield of \(\mathrm{N}_{2} \mathrm{O}_{5}\) when \(79.4 \mathrm{g}\) of \(\mathrm{P}_{4} \mathrm{O}_{10}\) are reacted with an excess of \(\mathrm{HNO}_{3}\) is \(60.3 \mathrm{g}\)

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Most popular questions from this chapter

At \(620 \mathrm{~K}\) the vapor density of ammonium chloride relative to hydrogen \(\left(\mathrm{H}_{2}\right)\) under the same conditions of temperature and pressure is \(14.5,\) although, according to its formula mass, it should have a vapor density of \(26.8 .\) How would you account for this discrepancy?

What is the change in oxidation number for the following reaction? $$3 \mathrm{O}_{2} \longrightarrow 2 \mathrm{O}_{3}$$

Starting with elemental phosphorus, \(\mathrm{P}_{4}\), show how you would prepare phosphoric acid.

Describe two reactions in which sulfuric acid acts as an oxidizing agent.

As mentioned in Chapter 3 , ammonium nitrate is the most important nitrogen- containing fertilizer in the world. Given only air and water as starting materials and any equipment and catalyst at your disposal, describe how you would prepare ammonium nitrate. State conditions under which you can increase the yield in each step.
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