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Chapter 4: Problem 136

A reaction mixture contains \(1.0 \mathrm{mol} \mathrm{CaCN}_{2}\) (calcium cyanamide) and \(1.0 \mathrm{mol} \mathrm{H}_{2} \mathrm{O}\). The maximum number of moles of \(\mathrm{NH}_{3}\) produced is (a) \(3.0 ;\) (b) 2.0 (c) between 1.0 and 2.0; (d) less than 1.0. $$\mathrm{CaCN}_{2}(\mathrm{s})+3 \mathrm{H}_{2} \mathrm{O}(\mathrm{l}) \longrightarrow \mathrm{CaCO}_{3}+2 \mathrm{NH}_{3}(\mathrm{g})$$

Short Answer

Expert verified
The maximum number of moles of \(\mathrm{NH}_{3}\) produced is less than 1.0, which is 'd'.

Step by step solution

01

Identify the limiting reagent

In this reaction mixture, it's given that \(1.0 \mathrm{mol} \mathrm{CaCN}_{2}\) and \(1.0 \mathrm{mol} \mathrm{H}_{2} \mathrm{O}\) are present. From the balanced chemical equation, it can be seen that one mole of \(\mathrm{CaCN}_{2}\) reacts with three moles of \(\mathrm{H}_{2} \mathrm{O}\). Since only one mole of \(\mathrm{H}_{2} \mathrm{O}\) is present, it will be used up before all of the \(\mathrm{CaCN}_{2}\) is, making \(\mathrm{H}_{2} \mathrm{O}\) the limiting reagent.
02

Apply stoichiometric ratios

Stoichiometric coefficients in the balanced chemical equation represent the molar ratios of reactants and products in the chemical reaction. In this reaction, two moles of \(\mathrm{NH}_{3}\) are produced for each mole of \(\mathrm{CaCN}_{2}\) and three moles of \(\mathrm{H}_{2} \mathrm{O}\).
03

Calculate the moles of product

Since the \(\mathrm{H}_{2} \mathrm{O}\) is the limiting reagent, its stoichiometric ratio with the \(\mathrm{NH}_{3}\) should be used to determine the amount of \(\mathrm{NH}_{3}\) that can be produced. Specifically, for every three moles of \(\mathrm{H}_{2} \mathrm{O}\), two moles of \(\mathrm{NH}_{3}\) are produced. Therefore, since we have only one mole of \(\mathrm{H}_{2} \mathrm{O}\), the ratio, 2/3, can be used to determine the moles of \(\mathrm{NH}_{3}\), which is \( \frac{2}{3} \times 1.0 \mathrm{mol} \) of \(\mathrm{NH}_{3}\).

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Key Concepts

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

Stoichiometry
Stoichiometry is like the mathematical backbone of chemistry. It involves the calculation of reactants and products in chemical reactions. You can think of it as a recipe, where you need specific amounts of ingredients (reactants) to get a certain amount of cake (product). In stoichiometry:
  • Balanced chemical equations provide the ratios of how much reactant is needed to produce a specific amount of product.
  • It uses the coefficients in chemical equations to relate moles of reactants to moles of products.
In our exercise, stoichiometry helps us understand how many moles of ammonia (\(\text{NH}_3\)) we can produce based on the moles of water (\(\text{H}_2\text{O}\)) since the water is the limiting reagent.
Chemical Reactions
A chemical reaction occurs when substances, called reactants, are transformed into new substances, called products. This transformation involves the breaking of old bonds and the formation of new ones. In the given equation,\(\mathrm{CaCN}_{2} + 3 \, \mathrm{H}_{2} \, \mathrm{O} \rightarrow \mathrm{CaCO}_{3} + 2 \, \mathrm{NH}_{3}\), we identify:
  • The reactants: calcium cyanamide (\(\text{CaCN}_2\)) and water (\(\text{H}_2\text{O}\)).
  • The products: calcium carbonate (\(\text{CaCO}_3\)) and ammonia (\(\text{NH}_3\)).
Knowing how chemical reactions proceed gives us the framework to predict and calculate outcomes, like how much of a product will form or which reactant will run out first.
Moles of Substances
The concept of moles is essential in chemistry. A mole is a unit of measurement that allows chemists to count particles like atoms and molecules in a bulk quantity. It is like a dozen, but instead of 12, one mole equals approximately 6.022 \( \times \) 10\(^{23}\) entities (Avogadro's number).
  • Moles help us relate measurements at the microscopic level to macroscopic quantities we can measure and observe.
  • They allow us to convert grams of a substance to the number of atoms or molecules and vice versa.
In our exercise, understanding moles allows us to translate 1.0 mole of water into a countable quantity that reacts with calcium cyanamide to produce ammonia. Remember, since 1 mole of \(\text{H}_2\text{O}\) is available and it acts as the limiting reagent, it will only partially allow the reaction to proceed, yielding less ammonia than if there were more water available.

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Most popular questions from this chapter

Nitrogen gas, \(\mathrm{N}_{2}\), can be prepared by passing gaseous ammonia over solid copper(II) oxide, \(\mathrm{CuO}\), at high temperatures. The other products of the reaction are solid copper, \(\mathrm{Cu},\) and water vapor. In a certain experiment, a reaction mixture containing \(18.1 \mathrm{g} \mathrm{NH}_{3}\) and \(90.4 \mathrm{g}\) CuO yields \(6.63 \mathrm{g} \mathrm{N}_{2}\). Calculate the percent yield for this experiment.

The following set of reactions is to be used as the basis of a method for producing nitric acid, \(\mathrm{HNO}_{3}\) Calculate the minimum masses of \(\mathrm{N}_{2}, \mathrm{H}_{2^{\prime}}\) and \(\mathrm{O}_{2}\) required per kilogram of \(\mathrm{HNO}_{3}\) $$\begin{array}{l} \mathrm{N}_{2}(\mathrm{g})+3 \mathrm{H}_{2}(\mathrm{g}) \longrightarrow 2 \mathrm{NH}_{3}(\mathrm{g}) \\ 4 \mathrm{NH}_{3}(\mathrm{g})+5 \mathrm{O}_{2}(\mathrm{g}) \longrightarrow 4 \mathrm{NO}(\mathrm{g})+6 \mathrm{H}_{2} \mathrm{O}(\mathrm{g}) \\ 2 \mathrm{NO}(\mathrm{g})+\mathrm{O}_{2}(\mathrm{g}) \longrightarrow 2 \mathrm{NO}_{2}(\mathrm{g}) \\ 3 \mathrm{NO}_{2}(\mathrm{g})+\mathrm{H}_{2} \mathrm{O}(\mathrm{l}) \longrightarrow 2 \mathrm{HNO}_{3}(\mathrm{aq})+\mathrm{NO}(\mathrm{g}) \end{array}$$

The minerals calcite, \(\mathrm{CaCO}_{3},\) magnesite, \(\mathrm{MgCO}_{3}\) and dolomite, \(\mathrm{CaCO}_{3} \cdot \mathrm{MgCO}_{3},\) decompose when strongly heated to form the corresponding metal oxide(s) and carbon dioxide gas. A 1.000 -g sample known to be one of the three minerals was strongly heated and \(0.477 \mathrm{g} \mathrm{CO}_{2}\) was obtained. Which of the three minerals was it?

A \(25.00 \mathrm{mL}\) sample of \(\mathrm{HCl}(\mathrm{aq})\) was added to a \(0.1000 \mathrm{g}\) sample of \(\mathrm{CaCO}_{3}\). All the \(\mathrm{CaCO}_{3}\) reacted, leaving some excess HCl(aq). \(\mathrm{CaCO}_{3}(\mathrm{s})+2 \mathrm{HCl}(\mathrm{aq}) \longrightarrow\) $$ \mathrm{CaCl}_{2}(\mathrm{aq})+\mathrm{H}_{2} \mathrm{O}(\mathrm{l})+\mathrm{CO}_{2}(\mathrm{g}) $$ The excess HCl(aq) required 43.82 mL of 0.01185 M \(\mathrm{Ba}(\mathrm{OH})_{2}\) to complete the following reaction. What was the molarity of the original HCl(aq)? $$2 \mathrm{HCl}(\mathrm{aq})+\mathrm{Ba}(\mathrm{OH})_{2}(\mathrm{aq}) \longrightarrow \mathrm{BaCl}_{2}(\mathrm{aq})+2 \mathrm{H}_{2} \mathrm{O}(\mathrm{l})$$

The incomplete combustion of gasoline produces \(\mathrm{CO}(\mathrm{g})\) as well as \(\mathrm{CO}_{2}(\mathrm{g}) .\) Write an equation for \((\mathrm{a})\) the complete combustion of the gasoline component octane, \(\mathrm{C}_{8} \mathrm{H}_{18}(\mathrm{l}),\) and \((\mathrm{b})\) incomplete combustion of octane with \(25 \%\) of the carbon appearing as \(\mathrm{CO}(\mathrm{g})\)
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