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Chapter 4: Problem 69

Ammonia can be generated by heating together the solids \(\mathrm{NH}_{4} \mathrm{Cl}\) and \(\mathrm{Ca}(\mathrm{OH})_{2} . \mathrm{CaCl}_{2}\) and \(\mathrm{H}_{2} \mathrm{O}\) are also formed. (a) If a mixture containing \(33.0 \mathrm{g}\) each of \(\mathrm{NH}_{4} \mathrm{Cl}\) and \(\mathrm{Ca}(\mathrm{OH})_{2}\) is heated, how many grams of \(\mathrm{NH}_{3}\) will form? (b) Which reactant remains in excess, and in what mass?

Short Answer

Expert verified
a) 10.5 g of \( NH_{3} \) will form. b) \( Ca(OH)_{2} \) remains in excess with a mass of 10.1 g.

Step by step solution

01

Balancing the Chemical Equation

The first step is to write down the chemical reaction and balance it: \[\mathrm{NH}_{4} \mathrm{Cl} + \mathrm{Ca}(\mathrm{OH})_{2} \rightarrow \mathrm{NH}_{3} + \mathrm{CaCl}_{2} + \mathrm{H}_{2} \mathrm{O}\] The balanced equation is: \[2\mathrm{NH}_{4} \mathrm{Cl} + \mathrm{Ca}(\mathrm{OH})_{2} \rightarrow 2\mathrm{NH}_{3} + \mathrm{CaCl}_{2} + 2\mathrm{H}_{2} \mathrm{O}\]
02

Calculation of Moles

The second step is to calculate the number of moles of reactants using the molar masses. For \(NH_{4}Cl\), the molar mass is 53.49 g/mol, and for \(Ca(OH)_{2}\), the molar mass is 74.09 g/mol. So, the moles of \( NH_{4}Cl \( are 33.0 g / 53.49 g/mol = 0.617 mol. For \( Ca(OH)_{2} \), the calculated moles are: 33.0 g / 74.09 g/mol = 0.445 mol.
03

Identification of Limiting Reactant

In the balanced chemical equation, the stoichiometric ratio between \( NH_{4}Cl \) and \( Ca(OH)_{2} \) is 2:1. This means that two moles of \( NH_{4}Cl \) react with each mole of \( Ca(OH)_{2} \). However, we have less than 2 times the amount of \( NH_{4}Cl \) than \( Ca(OH)_{2} \), so \( NH_{4}Cl \) will be entirely consumed first. Therefore, \( NH_{4}Cl \) is the limiting reactant.
04

Mass of Ammonia Formed

Based on the stoichiometry of the reaction, two moles of ammonia are produced for every two moles of \( NH_{4}Cl \) used. Since we start with 0.617 moles of \( NH_{4}Cl \), this means we form 0.617 moles of \( NH_{3} \). The molar mass of \( NH_{3} \) is 17.03 g/mol, so the mass of \( NH_{3} \) produced is 0.617 mol * 17.03 g/mol = 10.5 g.
05

Calculating Excess Reactant Mass

The stoichiometry also tells us that for every 2 moles of \( NH_{4}Cl \) that react, 1 mole of \( Ca(OH)_{2} \) is needed. Since we used all 0.617 moles of \( NH_{4}Cl \), this means we used 0.617 / 2 = 0.3085 mol of \( Ca(OH)_{2} \). If we started with 0.445 mol, this means we have 0.445 - 0.3085 = 0.1365 mol of \( Ca(OH)_{2} \) left. The molar mass of \( Ca(OH)_{2} \) is 74.09 g/mol, so the mass of \( Ca(OH)_{2} \) remaining is 0.1365 mol * 74.09 g/mol = 10.1 g.

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Most popular questions from this chapter

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