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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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Most popular questions from this chapter

A small piece of zinc is dissolved in \(50.00 \mathrm{mL}\) of \(1.035 \mathrm{M}\) HCl. At the conclusion of the reaction, the concentration of the \(50.00 \mathrm{mL}\) sample is redetermined and found to be \(0.812 \mathrm{M} \mathrm{HCl} .\) What must have been the mass of the piece of zinc that dissolved? $$\mathrm{Zn}(\mathrm{s})+2 \mathrm{HCl}(\mathrm{aq}) \longrightarrow \mathrm{ZnCl}_{2}(\mathrm{aq})+\mathrm{H}_{2}(\mathrm{g})$$

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