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Chapter 5: Problem 56

When \(35.6 \mathrm{~L}\) of ammonia and \(40.5 \mathrm{~L}\) of oxygen gas at \(\mathrm{STP}\) burn, nitrogen monoxide and water form. After the products return to STP, how many grams of nitrogen monoxide are present? $$\mathrm{NH}_{3}(g)+\mathrm{O}_{2}(g) \longrightarrow \mathrm{NO}(g)+\mathrm{H}_{2} \mathrm{O}(l) \quad[\text { unbalanced }]$$

Short Answer

Expert verified
43.52 g of \(\mathrm{NO}\)

Step by step solution

01

- Write the balanced equation

First, balance the chemical equation: \[ 4 \mathrm{NH}_3(g) + 5 \mathrm{O}_2(g) \rightarrow 4 \mathrm{NO}(g) + 6 \mathrm{H}_2 \mathrm{O}(l) \] This shows the molar ratios of reactants and products.
02

- Convert gas volumes to moles (STP)

At STP (Standard Temperature and Pressure), 1 mole of any gas occupies 22.4 L. Calculate moles of \(\mathrm{NH}_3\): \[ n(\mathrm{NH}_3) = \frac{35.6 \mathrm{~L}}{22.4 \mathrm{~L/mol}} = 1.59 \text{ moles} \] Calculate moles of \(\mathrm{O}_2\): \[ n(\mathrm{O}_2) = \frac{40.5 \mathrm{~L}}{22.4 \mathrm{~L/mol}} = 1.81 \text{ moles} \]
03

- Determine the limiting reactant

Compare mole ratios with the balanced equation. From the balanced equation, 4 moles of \(\mathrm{NH}_3\) react with 5 moles of \(\mathrm{O}_2\). Calculate the required amount of \(\mathrm{O}_2\) for the given \(\mathrm{NH}_3\): \[ \frac{5}{4} \times 1.59 \text{ moles} \mathrm{NH}_3 = 1.99 \text{ moles} \mathrm{O}_2\] Since we only have 1.81 moles of \(\mathrm{O}_2\), \(\mathrm{O}_2\) is the limiting reactant.
04

- Calculate moles of \(\mathrm{NO}\) formed

Using the mole ratio from the balanced equation (5 moles \(\mathrm{O}_2\) yields 4 moles \(\mathrm{NO}\)): \[ n(\mathrm{NO}) = \frac{4}{5} \times 1.81 \text{ moles} \mathrm{O}_2 = 1.45 \text{ moles} \mathrm{NO} \]
05

- Convert moles of \(\mathrm{NO}\) to grams

Find the molar mass of \(\mathrm{NO}\): \[ M(\mathrm{NO}) = 14.01 + 16.00 = 30.01 \text{ g/mol} \] Calculate the mass of \(\mathrm{NO}\): \[ \text{mass} = 1.45 \text{ moles} \times 30.01 \text{ g/mol} = 43.52 \text{ g} \]

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Key Concepts

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

Limiting Reactant
The concept of the limiting reactant is key in solving stoichiometry problems. In a chemical reaction, the limiting reactant is the substance that is completely consumed first. This limits the amount of products that can be formed. To determine the limiting reactant, you must compare the mole ratios of the reactants used in the experiment to the ratios in the balanced chemical equation. In our exercise example, we calculated the moles of \(\mathrm{NH}_3\) and \(\mathrm{O}_2\) to determine which one runs out first. By comparing the moles needed according to the balanced equation, we found out that \(\mathrm{O}_2\) is the limiting reactant.
Gas Laws at STP
STP stands for Standard Temperature and Pressure, which is defined as 0掳C (273.15 K) and 1 atm pressure. Under these conditions, one mole of any ideal gas occupies 22.4 liters. This relationship helps simplify the conversion between volume and moles for gas calculations in stoichiometry. In our problem, we used this property to convert the volumes of ammonia (\(\mathrm{NH}_3\)) and oxygen (\(\mathrm{O}_2\)) into moles by dividing their volumes by 22.4 L/mol. This is an essential step in calculations involving gases.
Balanced Chemical Equation
Balancing a chemical equation is necessary to accurately represent the reaction's stoichiometry. It ensures that the number of atoms for each element is the same on both sides of the equation, following the law of conservation of mass. In the given problem, the unbalanced equation was: \[\mathrm{NH}_{3}(g) + \mathrm{O}_{2}(g) \rightarrow \mathrm{NO}(g) + \mathrm{H}_2 \mathrm{O}(l) \] By balancing, we obtained: \[ 4\mathrm{NH}_3(g) + 5\mathrm{O}_2(g) \rightarrow 4\mathrm{NO}(g) + 6\mathrm{H}_2 \mathrm{O}(l) \] This shows the correct molar ratios needed to calculate the moles of products formed.
Molar Mass Calculation
Molar mass is critical for converting moles to grams and vice versa. It is the mass of one mole of a substance, usually expressed in grams per mole (g/mol). To find the molar mass, sum the atomic masses of all the atoms in the molecule. For example, the molar mass of nitrogen monoxide (\(\mathrm{NO}\)) is calculated as follows: \[ M(\mathrm{NO}) = 14.01 \text{ (N)} + 16.00 \text{ (O)} = 30.01 \text{ g/mol} \] This value is then used to convert moles of \(\mathrm{NO}\) to grams, as shown in our exercise.
Chemical Reactions
Chemical reactions involve the transformation of reactants into products through the breaking and forming of bonds. Understanding how reactants interact and the stoichiometric relationships between them allows us to predict the amounts of products formed. In our example, ammonia (\(\mathrm{NH}_3\)) reacts with oxygen (\(\mathrm{O}_2\)) to form nitrogen monoxide (\(\mathrm{NO}\)) and water (\(\mathrm{H}_2\mathrm{O}\)). By balancing the equation and performing calculations based on the limiting reactant, we determined the amount of nitrogen monoxide produced. Knowing these principles helps in many areas of chemistry, from laboratory experiments to industrial processes.

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Most popular questions from this chapter

Convert the following: (a) 0.745 atm to \(\mathrm{mmHg}\) (b) 992 torr to bar (c) \(365 \mathrm{kPa}\) to atm (d) \(804 \mathrm{mmHg}\) to \(\mathrm{kPa}\)

Combustible vapor-air mixtures are flammable over a limited range of concentrations. The minimum volume \(\%\) of vapor that gives a combustible mixture is called the lower flammable limit (LFL). Generally, the LFL is about half the stoichiometric mixture, which is the concentration required for complete combustion of the vapor in air. (a) If oxygen is 20.9 vol \% of air, estimate the LFL for \(n\) -hexane, \(\mathrm{C}_{6} \mathrm{H}_{14}\). (b) What volume (in \(\mathrm{mL}\) ) of \(n\) -hexane \(\left(d=0.660 \mathrm{~g} / \mathrm{cm}^{3}\right)\) is required to produce a flammable mixture of hexane in \(1.000 \mathrm{~m}^{3}\) of air at \(\mathrm{STP} ?\)

The Hawaian volcano Kilauea emits an average of \(1.5 \times 10^{3} \mathrm{~m}^{3}\) of gas each day, when corrected to \(298 \mathrm{~K}\) and \(1.00 \mathrm{~atm} .\) The mixture contains gases that contribute to global warming and acid rain, and some are toxic. An atmospheric chemist analyzes a sample and finds the following mole fractions: \(0.4896 \mathrm{CO}_{2}, 0.0146 \mathrm{CO}, 0.3710 \mathrm{H}_{2} \mathrm{O}\) \(0.1185 \mathrm{SO}_{2}, 0.0003 \mathrm{~S}_{2}, 0.0047 \mathrm{H}_{2}, 0.0008 \mathrm{HCl},\) and \(0.0003 \mathrm{H}_{2} \mathrm{~S} .\) How many metric tons (t) of each gas are emitted per year ( \(1 \mathrm{t}=1000 \mathrm{~kg}\) )?

A large portion of metabolic energy arises from the biological combustion of glucose: $$\mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}(s)+6 \mathrm{O}_{2}(g) \longrightarrow 6 \mathrm{CO}_{2}(g)+6 \mathrm{H}_{2} \mathrm{O}(g)$$ (a) If this reaction is carried out in an expandable container at \(37^{\circ} \mathrm{C}\) and \(780 .\) torr, what volume of \(\mathrm{CO}_{2}\) is produced from \(20.0 \mathrm{~g}\) of glucose and excess \(\mathrm{O}_{2} ?\) (b) If the reaction is carried out at the same conditions with the stoichiometric amount of \(\mathrm{O}_{2}\), what is the partial pressure of each gas when the reaction is \(50 \%\) complete \((10.0 \mathrm{~g}\) of glucose remains)?

5.102 In a bromine-producing plant, how many liters of gaseous elemental bromine at \(300^{\circ} \mathrm{C}\) and 0.855 atm are formed by the reaction of \(275 \mathrm{~g}\) of sodium bromide and \(175.6 \mathrm{~g}\) of sodium bromate in aqueous acid solution? (Assume that no Br \(_{2}\) dissolves.) $$\begin{array}{r}5 \mathrm{NaBr}(a q)+\mathrm{NaBrO}_{3}(a q)+3 \mathrm{H}_{2} \mathrm{SO}_{4}(a q) \longrightarrow \\\3 \mathrm{Br}_{2}(g)+3 \mathrm{Na}_{2} \mathrm{SO}_{4}(a q)+3 \mathrm{H}_{2} \mathrm{O}(g)\end{array}$$
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