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Chapter 4: Problem 122

For a specific reaction, ammonium dichromate is the only reactant and chromium(III) oxide and water are two of the three products. What is the third product and how many grams of this product are produced per kilogram of ammonium dichromate decomposed?

Short Answer

Expert verified
The third product is Nitrogen gas (\( N_2 \)). Approximately 111.17 g of this product are produced per kilogram of ammonium dichromate decomposed.

Step by step solution

01

Identify the Balanced Chemical Reaction

Firstly, the balanced chemical reaction for the decomposition of ammonium dichromate (\( (NH_4)_2Cr_2O_7 \)) is: \[ (NH_4)_2Cr_2O_7 \rightarrow Cr_2O_3 + 4 H_2O + N_2 \] So, the third product is Nitrogen gas (\( N_2 \))
02

Calculate Molar Masses

Next, calculate the molar masses of the reactant and the third product: \[ Molar\ mass\ of\ (NH_4)_2Cr_2O_7 = 252.07\ g/mol \] \[ Molar\ mass\ of\ N_2 = 28.014\ g/mol \]
03

Calculate the Amount of Third Product Produced

The stoichiometry of the reaction tells us that 1 mole of \( (NH_4)_2Cr_2O_7 \) decomposed yields 1 mole of \( N_2 \). So, if we have 1000 g (1 kg) of \( (NH_4)_2Cr_2O_7 \), this corresponds to \( \frac{1000}{252.07} \) moles. Hence, the same number of moles of \( N_2 \) is produced, which corresponds to \( \frac{1000}{252.07} \times 28.014 = 111.17 \) g of \( N_2 \).

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Key Concepts

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

Balanced Chemical Equation
In a chemical reaction, it's crucial to have a balanced chemical equation. This ensures the law of conservation of mass is satisfied, which states that mass cannot be created or destroyed in a chemical reaction. All atoms present in the reactants must be accounted for in the products. In our example, the decomposition of ammonium dichromate is balanced as follows: \[(NH_4)_2Cr_2O_7 \rightarrow Cr_2O_3 + 4 H_2O + N_2\] Examining this equation, you can see that the atoms on both sides are equal. Each nitrogen, hydrogen, chromium, and oxygen atom in the reactant matches with the respective atoms in the products. The third product formed is nitrogen gas \(N_2\). To determine this, you must balance and count the elements of each product to ensure the total count remains the same.
Stoichiometry
Stoichiometry is the part of chemistry that deals with the relative quantities of reactants and products in chemical reactions. It uses the coefficients from the balanced equation to make calculations about the masses of reactants and products. Let's apply stoichiometry to our decomposition reaction: For every mole of \((NH_4)_2Cr_2O_7\) that decomposes, one mole of each product is formed, including nitrogen gas \(N_2\). This mole-to-mole ratio is crucial for calculating how much of a substance is produced or used up in a reaction. By using these ratios derived from the balanced equation, you can determine quantities involved in the reactions, such as in our case, calculating the grams of nitrogen gas produced from a certain amount of ammonium dichromate.
Molar Mass Calculation
Molar mass is a fundamental concept for connecting the mass of a substance to the amount in moles. It is the mass of one mole of a given substance (atoms, molecules, etc.) and is usually expressed in grams per mole \(g/mol\). Calculating molar mass correctly enables precise stoichiometric calculations. For the decomposition reaction of ammonium dichromate:
  • The molar mass of \((NH_4)_2Cr_2O_7\) is calculated by adding the atomic masses of all its constituent atoms: \(252.07\ g/mol\).
  • The molar mass of nitrogen gas \(N_2\) is the sum of the masses of two nitrogen atoms: \(28.014\ g/mol\).
By determining these molar masses, you can transform your stoichiometric calculations from moles to grams, which are more practical for real-world applications. This allows you to conclude that 1000 grams of ammonium dichromate (H_2O\ldots) will produce 111.17 grams of nitrogen gas by using these calculations and the earlier computed stoichiometric ratios.

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Most popular questions from this chapter

Silver nitrate is a very expensive chemical. For a particular experiment, you need \(100.0 \mathrm{mL}\) of \(0.0750 \mathrm{M}\) \(\mathrm{AgNO}_{3},\) but only \(60 \mathrm{mL}\) of \(0.0500 \mathrm{M} \mathrm{AgNO}_{3}\) is available. You decide to pipet exactly \(50.00 \mathrm{mL}\) of the solution into a \(100.0 \mathrm{mL}\) flask, add an appropriate mass of \(\mathrm{AgNO}_{3},\) and then dilute the resulting solution to exactly \(100.0 \mathrm{mL}\). What mass of \(\mathrm{AgNO}_{3}\) must you use?

An organic liquid is either methyl alcohol \(\left(\mathrm{CH}_{3} \mathrm{OH}\right)\) ethyl alcohol \(\left(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{OH}\right),\) or a mixture of the two. A 0.220-g sample of the liquid is burned in an excess of \(\mathrm{O}_{2}(\mathrm{g})\) and yields \(0.352 \mathrm{g} \mathrm{CO}_{2}(\mathrm{g}) .\) Is the liquid a pure alcohol or a mixture of the two?

When water and methanol, \(\mathrm{CH}_{3} \mathrm{OH}(\mathrm{l}),\) are mixed, the total volume of the resulting solution is not equal to the sum of the pure liquid volumes. (Refer to Exercise 99 for an explanation.) When \(72.061 \mathrm{g} \mathrm{H}_{2} \mathrm{O}\) and \(192.25 \mathrm{g}\) \(\mathrm{CH}_{3} \mathrm{OH}\) are mixed at \(25^{\circ} \mathrm{C},\) the resulting solution has a density of \(0.86070 \mathrm{g} / \mathrm{mL} .\) At \(25^{\circ} \mathrm{C},\) the densities of water and methanol are \(0.99705 \mathrm{g} / \mathrm{mL}\) and \(0.78706\) \(\mathrm{g} / \mathrm{mL},\) respectively. (a) Calculate the volumes of the pure liquid samples and the solution, and show that the pure liquid volumes are not additive. [ Hint: Although the volumes are not additive, the masses are.] (b) Calculate the molarity of \(\mathrm{CH}_{3} \mathrm{OH}\) in this solution.

A method for eliminating oxides of nitrogen (e.g., \(\mathrm{NO}_{2}\) ) from automobile exhaust gases is to pass the exhaust gases over solid cyanuric acid, \(\mathrm{C}_{3} \mathrm{N}_{3}(\mathrm{OH})_{3}\) When the hot exhaust gases come in contact with cyanuric acid, solid \(\mathrm{C}_{3} \mathrm{N}_{3}(\mathrm{OH})_{3}\) decomposes into isocyanic acid vapor, HNCO(g), which then reacts with \(\mathrm{NO}_{2}\) in the exhaust gases to give \(\mathrm{N}_{2}, \mathrm{CO}_{2^{\prime}}\) and \(\mathrm{H}_{2} \mathrm{O}\) How many grams of \(\mathrm{C}_{3} \mathrm{N}_{3}(\mathrm{OH})_{3}\) are needed per gram of \(\mathrm{NO}_{2}\) in this method? [Hint: To balance the equation for reaction between HNCO and \(\mathrm{NO}_{2}\), balance with respect to each kind of atom in this order: \(\mathrm{H}, \mathrm{C}, \mathrm{O}, \text { and } \mathrm{N} .]\)

Suppose that reactions (a) and (b) each have a \(92 \%\) yield. Starting with \(112 \mathrm{g} \mathrm{CH}_{4}\) in reaction \((\mathrm{a})\) and an excess of \(\mathrm{Cl}_{2}(\mathrm{g}),\) how many grams of \(\mathrm{CH}_{2} \mathrm{Cl}_{2}\) are formed in reaction (b)? (a) \(\mathrm{CH}_{4}+\mathrm{Cl}_{2} \longrightarrow \mathrm{CH}_{3} \mathrm{Cl}+\mathrm{HCl}\) (b) \(\mathrm{CH}_{3} \mathrm{Cl}+\mathrm{Cl}_{2} \longrightarrow \mathrm{CH}_{2} \mathrm{Cl}_{2}+\mathrm{HCl}\)
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