The percentage yield of nitric acid is 95\(\% .\) If 9.88 \(\mathrm{kg}\) of nitrogen dioxide react, what mass of nitric acid is isolated? $$3 \mathrm{NO}_{2}(g)+\mathrm{H}_{2} \mathrm{O}(g) \rightarrow 2 \mathrm{HNO}_{3}(a q)+\mathrm{NO}(g)$$

Short Answer

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Perform the calculations based on the steps described and you should find the mass of nitric acid isolated.

Step by step solution

01

Calculate Molar Masses

Firstly, let's find the molar mass of nitrogen dioxide \(\mathrm{NO}_{2}\) and nitric acid \(\mathrm{HNO}_{3}\). For \(\mathrm{NO}_{2}\), the molar mass is \(14.01 \, (N) + 32.00 \, (2O) = 46.01 \, \mathrm{g/mol}\), and for \(\mathrm{HNO}_{3}\), it's \(1.01 \, g/mol (H) + 14.01 \, g/mol (N) + 48 \, g/mol (3O) = 63.02 \, g/mol\).
02

Calculate moles of nitrogen dioxide

Now, let's calculate the moles of nitrogen dioxide that react. We have 9.88 kg of \(\mathrm{NO}_{2}\) which is 9880 g. Therefore, moles of \(\mathrm{NO}_{2} = \frac{mass}{molar \, mass} = \frac{9880}{46.01} \, mol\).
03

Use stoichiometry to calculate moles of nitric acid

From the chemical reaction, we know that 3 moles of nitrogen dioxide produce 2 moles of nitric acid. Using the stoichiometric ratio, we can calculate the theoretical moles of nitric acid produced. \( \text{Moles of } \mathrm{HNO}_{3} = \text{Moles of } \mathrm{NO}_{2} \times \left(\frac{2 \, \text{moles of } \mathrm{HNO}_{3}}{3 \, \text{moles of }\mathrm{NO}_{2}}\right)\).
04

Determine theoretical mass of nitric acid

We can calculate the theoretical mass of nitric acid that could be produced by multiplying the moles of nitric acid by its molar mass. \( \text{Mass of } \mathrm{HNO}_{3} = \text{moles of } \mathrm{HNO}_{3} \times \text{molar mass of } \mathrm{HNO}_{3} \).
05

Calculate actual mass of nitric acid isolated

Finally, the actual yield of nitric acid is 95% of the theoretical yield. Thus, the mass of nitric acid isolated = 0.95 (95%) times the theoretical mass calculated in the previous step.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Percentage Yield Calculation
When performing chemical reactions, the percentage yield is a measure of the efficiency of a reaction. It is calculated by taking the actual yield (the amount of product obtained) and dividing it by the theoretical yield (the amount of product that could be formed based on stoichiometry), then multiplying the result by 100 to get a percentage.

For example, if the theoretical yield of a reaction is 10 grams but only 9 grams of product were actually isolated, the percentage yield would be \( \frac{9 g}{10 g} \times 100\% = 90\% \).

The percentage yield can be influenced by factors such as incomplete reactions, impurities, or losses during product recovery. Thus, it's rare to achieve a 100% yield in reality. Knowing the percentage yield helps in evaluating the reaction's practicality and cost-efficiency.
Molar Mass Determination
The molar mass of a compound is the mass of one mole of its molecules and is measured in grams per mole (g/mol). To determine the molar mass, we add together the atomic masses of each element within the compound according to its chemical formula.

For instance, to find the molar mass of nitric acid \( \mathrm{HNO}_3 \), we sum the masses of hydrogen \(1.01 \, g/mol\), nitrogen \(14.01 \, g/mol\), and oxygen \(16.00 \, g/mol\, multiplied by 3 since there are three oxygen atoms\), resulting in a molar mass of \(63.02\, g/mol\).

Being able to determine molar mass is fundamental in stoichiometry as it allows for conversion between mass and moles, crucial for quantitatively analyzing chemical reactions.
Chemical Stoichiometry
Chemical stoichiometry involves the quantitative relationship between reactants and products in a chemical reaction. It helps predict how much product will form from a given amount of reactant or how much reactant is needed to make a desired amount of product.

Stoichiometry is grounded on the law of conservation of mass and the balanced chemical equation. For example, in the reaction \(3 \mathrm{NO}_2(g) + \mathrm{H}_2\mathrm{O}(g) \rightarrow 2\mathrm{HNO}_3(aq) + \mathrm{NO}(g)\), the ratio of moles of nitrogen dioxide to moles of nitric acid is 3:2. This ratio means that every three moles of \(\mathrm{NO}_2\) should theoretically yield two moles of \(\mathrm{HNO}_3\).

By using these ratios, we can perform calculations to scale up or down the amounts needed for a reaction or to predict the yield from a given quantity of starting material.
Theoretical Yield Calculation
The theoretical yield is the maximum amount of product that could be produced in a chemical reaction if everything reacted completely. It's calculated based on the reaction's balanced equation and the limiting reactant, which is the reactant that will run out first, limiting the amount of product formed.

To calculate the theoretical yield, one must first identify the mole ratio of the reactant to the product from the balanced equation. Then, determine the amount of moles of the reactant being used. Multiply the moles of the reactant by the mole ratio to find the moles of product that should form theoretically. Finally, convert this to mass using the molar mass of the product.

In practice, the actual yield will often be less than the theoretical yield due to various factors such as side reactions, incomplete reactions, or loss of product during purification.

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Most popular questions from this chapter

Use the equation provided to answer the questions that follow. $$2 \mathrm{Na}+2 \mathrm{H}_{2} \mathrm{O} \longrightarrow 2 \mathrm{NaOH}+\mathrm{H}_{2}$$ \begin{equation}\begin{array}{l}{\text { a. How many molecules of } \mathrm{H}_{2} \text { could be }} \\ {\text { made from } 27.6 \mathrm{g} \mathrm{H}_{2} \mathrm{O} \text { ? }} \\ {\text { b. How many atoms of Na will completely }} \\ {\text { react with } 12.9 \mathrm{g} \mathrm{H}_{2} \mathrm{O} \text { ? }} \\ {\text { c. How many molecules of } \mathrm{H}_{2} \text { could form }} \\ {\text { when } 6.59 \times 10^{20} \text { atoms Na react? }}\end{array}\end{equation}

The chemical equation for the formation of water is $$2 \mathrm{H}_{2}+\mathrm{O}_{2} \rightarrow 2 \mathrm{H}_{2} \mathrm{O}$$ \begin{equation}\begin{array}{l}{\text { a. If } 3.3 \mathrm{mol} \mathrm{O}_{2} \text { are used, how many moles }} \\ {\text { of } \mathrm{H}_{2} \text { are needed? }} \\ {\text { b. How many moles } \mathrm{O}_{2} \text { must react with }} \\ {\text { excess } \mathrm{H}_{2} \text { to form } 6.72 \mathrm{mol} \mathrm{H}_{2} \mathrm{O} \text { ? }} \\\ {\text { c. If you wanted to make } 8.12 \mathrm{mol} \mathrm{H}_{2} \mathrm{O},} \\ {\text { how many moles of } \mathrm{H}_{2} \text { would you need? }}\end{array}\end{equation}

Use stoichiometry to explain the following problems that a lawn mower may have. a. A lawn mower fails to start because the engine floods. b. A lawn mower stalls after starting cold and idling

Describe the relationship between the limiting reactant and the theoretical yield.

Use the balanced equation below to write mole ratios for the situations that follow. $$2 \mathrm{H}_{2}(g)+\mathrm{O}_{2}(\mathrm{g}) \rightarrow 2 \mathrm{H}_{2} \mathrm{O}(g)$$ \begin{equation}\begin{array}{l}{\text { a. calculating mol } \mathrm{H}_{2} \mathrm{O} \text { given mol } \mathrm{H}_{2}} \\ {\text { b. calculating mol } \mathrm{O}_{2} \text { given mol } \mathrm{H}_{2} \mathrm{O}} \\ {\text { c. calculating mol } \mathrm{H}_{2} \text { given mol } \mathrm{O}_{2}}\end{array}\end{equation}

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