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Chapter 9: Problem 37

Ammonium nitrate has been used as a high explosive because it is unstable and decomposes into several gaseous substances. The rapid expansion of the gaseous substances produces the explosive force. $$\mathrm{NH}_{4} \mathrm{NO}_{3}(s) \rightarrow \mathrm{N}_{2}(g)+\mathrm{O}_{2}(g)+\mathrm{H}_{2} \mathrm{O}(g)$$ Calculate the mass of each product gas if \(1.25 \mathrm{g}\) of ammonium nitrate reacts.

Short Answer

Expert verified
When 1.25g of ammonium nitrate reacts, the mass of each gaseous product formed is: \(N_2: 0.4363g\), \(O_2: 0.4992g\), and \(H_2O: 0.2811g\).

Step by step solution

01

Calculate the molar mass of ammonium nitrate and all products

We need to find the molar masses of the reactant (NH鈧凬O鈧�) and products (N鈧�, O鈧�, and H鈧侽) to convert mass to moles and vice versa. NH鈧凬O鈧�: (1 x 14.01) + (4 x 1.01) + (1 x 14.01) + (3 x 16) = 80.05 g/mol N鈧�: 2 x 14.01 = 28.02 g/mol O鈧�: 2 x 16 = 32 g/mol H鈧侽: (2 x 1.01) + 16 = 18.02 g/mol
02

Calculate the moles of ammonium nitrate

To find the moles of ammonium nitrate, we will use the given mass (1.25 g) and divide it by the molar mass (80.05 g/mol) calculated in Step 1. Moles of NH鈧凬O鈧� = (1.25 g) / (80.05 g/mol) = 0.0156 mol
03

Use stoichiometry to find the moles of each product

The stoichiometry of the balanced chemical equation shows a 1:1 ratio between the reactants and each product. Therefore, the moles of each product are equal to the moles of the reactant. Moles of N鈧� = 0.0156 mol Moles of O鈧� = 0.0156 mol Moles of H鈧侽 = 0.0156 mol
04

Calculate the mass of each product

Using the moles of each product and their respective molar masses obtained in Step 1, we can now calculate the mass of each gaseous product. Mass of N鈧� = (0.0156 mol) x (28.02 g/mol) = 0.4363 g Mass of O鈧� = (0.0156 mol) x (32 g/mol) = 0.4992 g Mass of H鈧侽 = (0.0156 mol) x (18.02 g/mol) = 0.2811 g Thus, the mass of each gaseous product formed when 1.25g of ammonium nitrate reacts is: - N鈧�: 0.4363 g - O鈧�: 0.4992 g - H鈧侽: 0.2811 g

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Key Concepts

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

Molar Mass Calculation
Molar mass is a fundamental concept in chemistry that relates to the mass of one mole of a substance. It allows you to convert between the mass of a compound and the number of moles, which represents the number of molecules or atoms present.
  • The molar mass of a compound is determined by summing up the atomic masses of its constituent elements, based on their presence in the chemical formula.
  • For instance, in ammonium nitrate (NH鈧凬O鈧�), the molar mass is calculated by adding the atomic masses of nitrogen, hydrogen, and oxygen in the compound:
    • Nitrogen contributes two atoms each with an atomic mass of 14.01,
    • Hydrogen contributes four atoms each with an atomic mass of 1.01,
    • Oxygen contributes three atoms each with an atomic mass of 16.
    So, the total molar mass = (1 x 14.01) + (4 x 1.01) + (1 x 14.01) + (3 x 16) = 80.05 g/mol.
  • This calculation is essential as it lays the groundwork for converting the mass of the compound into moles, a process necessary for stoichiometric calculations.
Stoichiometry
Stoichiometry involves understanding and using the quantitative relationships in chemical reactions. In essence, it helps you predict the amounts of substances consumed or produced.
  • To determine this, one must first write a balanced chemical equation, which ensures mass conservation of each element.
  • A crucial aspect is the mole ratio, which is derived from the coefficients of the substances in the balanced equation. In the decomposition of ammonium nitrate to produce N鈧�, O鈧�, and H鈧侽, there's a stoichiometric ratio of 1:1, indicating that one mole of NH鈧凬O鈧� yields one mole of each product.
  • This concept allows you to determine how the amount of reactant relates to the amount of products. For instance, if you know the moles of NH鈧凬O鈧�, you directly know the moles of each product as they have a 1:1 relationship.
Chemical Reaction Decomposition
Chemical decomposition is a type of chemical reaction where one compound breaks down into two or more simpler substances. Ammonium nitrate, for example, decomposes to form nitrogen, oxygen, and water vapor.
  • Decomposition reactions often require energy, like heat, to break chemical bonds in the compound. In the case of ammonium nitrate, this results in gaseous products that expand rapidly, demonstrating explosive characteristics.
  • Such a reaction can be represented by a simple equation: NH鈧凬O鈧�(s) 鈫� N鈧�(g) + O鈧�(g) + H鈧侽(g). This explains the transformation of a solid reactant into gaseous products, illustrating conservation of mass where the mass of reactants equals the mass of products in terms of both quantity and type of atoms.
  • This particular decomposition is important industrially and historically due to its explosive potential, highlighting the significance of understanding chemical processes both in educational settings and practical applications.

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Most popular questions from this chapter

Bottled propane is used in areas away from natural gas pipelines for cooking and heating, and is also the source of heat in most gas barbecue grills. Propane burns in oxygen according to the following balanced chemical equation: $$\mathrm{C}_{3} \mathrm{H}_{8}(g)+5 \mathrm{O}_{2}(g) \rightarrow 3 \mathrm{CO}_{2}(g)+4 \mathrm{H}_{2} \mathrm{O}(g)$$ Calculate the mass in grams of water vapor produced if 3.11 mol of propane is burned.

When elemental copper is strongly heated with sulfur, a mixture of CuS and \(\mathrm{Cu}_{2} \mathrm{S}\) is produced, with CuS predominating. $$\begin{array}{c} \mathrm{Cu}(s)+\mathrm{S}(s) \rightarrow \mathrm{CuS}(s) \\ 2 \mathrm{Cu}(s)+\mathrm{S}(s) \rightarrow \mathrm{Cu}_{2} \mathrm{S}(s) \end{array}$$ What is the theoretical yield of CuS when 31.8 g of \(\mathrm{Cu}(s)\) is heated with \(50.0 \mathrm{g}\) of \(\mathrm{S} ?\) (Assume only CuS is produced in the reaction.) What is the percent yield of CuS if only \(40.0 \mathrm{g}\) of CuS can be isolated from the mixture?

Consider the balanced equation $$ \mathrm{C}_{3} \mathrm{H}_{8}(g)+5 \mathrm{O}_{2}(g) \rightarrow 3 \mathrm{CO}_{2}(g)+4 \mathrm{H}_{2} \mathrm{O}(g) $$ What mole ratio enables you to calculate the number of moles of oxygen needed to react exactly with a given number of moles of \(\mathrm{C}_{3} \mathrm{H}_{8}(g) ?\) What mole ratios enable you to calculate how many moles of each product form from a given number of moles of \(\mathrm{C}_{3} \mathrm{H}_{8} ?\)

When the sugar glucose, \(\mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6},\) is burned in air, carbon dioxide and water vapor are produced. Write the balanced chemical equation for this process, and calculate the theoretical yield of carbon dioxide when \(1.00 \mathrm{g}\) of glucose is burned completely.

For each of the following unbalanced chemical equations, suppose \(1.00 \mathrm{g}\) of each reactant is taken. Show by calculation which reactant is limiting. Calculate the mass of each product that is expected. a. \(\mathrm{UO}_{2}(s)+\mathrm{HF}(a q) \rightarrow \mathrm{UF}_{4}(a q)+\mathrm{H}_{2} \mathrm{O}(l)\) b. \(\mathrm{NaNO}_{3}(a q)+\mathrm{H}_{2} \mathrm{SO}_{4}(a q) \rightarrow\) \(\mathrm{Na}_{2} \mathrm{SO}_{4}(a q)+\mathrm{HNO}_{3}(a q)\) c. \(\mathrm{Zn}(s)+\mathrm{HCl}(a q) \rightarrow \mathrm{ZnCl}_{2}(a q)+\mathrm{H}_{2}(g)\) d. \(\mathrm{B}(\mathrm{OH})_{3}(s)+\mathrm{CH}_{3} \mathrm{OH}(l) \rightarrow \mathrm{B}\left(\mathrm{OCH}_{3}\right)_{3}(s)+\mathrm{H}_{2} \mathrm{O}(l)\)
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