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Chapter 3: Problem 67

Ammonia is a principal nitrogen fertilizer. It is prepared by the reaction between hydrogen and nitrogen. $$ 3 \mathrm{H}_{2}(g)+\mathrm{N}_{2}(g) \longrightarrow 2 \mathrm{NH}_{3}(g) $$ In a particular reaction, 6.0 moles of \(\mathrm{NH}_{3}\) were produced. How many moles of \(\mathrm{H}_{2}\) and how many moles of \(\mathrm{N}_{2}\) were reacted to produce this amount of \(\mathrm{NH}_{3} ?\)

Short Answer

Expert verified
To produce 6 moles of \(NH_{3}\), 9 moles of \(H_{2}\) and 3 moles of \(N_{2}\) were reacted.

Step by step solution

01

Identify the given and the needed information

In this question, it is given that the number of moles of ammonia (NH3) produced is 6.0 moles. We need to determine the number of moles of hydrogen (H2) and nitrogen (N2) that reacted to form this amount of NH3. The chemical equation provided \(3H_{2}(g) + N_{2}(g)\rightarrow 2NH_{3}(g)\) helps us establish the ratio of the reactants to the product.
02

Use stoichiometric ratios to calculate moles of H2

From the balanced chemical equation, we know that 3 moles of H2 react to form 2 moles of NH3. Thus, the ratio of H2 to NH3 is \(\frac{3}{2}\). Therefore, the number of moles of H2 is calculated by multiplying the ratio \(\frac{3}{2}\) by the number of moles of NH3, which is 6 moles. So, \(moles \, of \, H_{2} = \frac{3}{2} \times 6 = 9 \, moles\)
03

Use stoichiometric ratios to calculate moles of N2

From the balanced chemical equation, we know that 1 mole of N2 reacts to form 2 moles of NH3. Thus, the ratio of N2 to NH3 is \(\frac{1}{2}\). Therefore, the number of moles of N2 is calculated by multiplying the ratio \(\frac{1}{2}\) by the number of moles of NH3, which is 6 moles. So, \(moles \, of \, N_{2} = \frac{1}{2} \times 6 = 3 \, moles\)

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Most popular questions from this chapter

Why is the theoretical yield of a reaction determined only by the amount of the limiting reagent?

The atomic masses of \({ }_{3}^{6} \mathrm{Li}\) and \({ }_{3}^{7} \mathrm{Li}\) are \(6.0151 \mathrm{amu}\) and 7.0160 amu, respectively. Calculate the natural abundances of these two isotopes. The average atomic mass of Li is 6.941 amu.

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The depletion of ozone \(\left(\mathrm{O}_{3}\right)\) in the stratosphere has been a matter of great concern among scientists in recent years. It is believed that ozone can react with nitric oxide (NO) that is discharged from the high-altitude jet plane, the SST. The reaction is $$ \mathrm{O}_{3}+\mathrm{NO} \longrightarrow \mathrm{O}_{2}+\mathrm{NO}_{2} $$ If \(0.740 \mathrm{~g}\) of \(\mathrm{O}_{3}\) reacts with \(0.670 \mathrm{~g}\) of \(\mathrm{NO}\), how many grams of \(\mathrm{NO}_{2}\) will be produced? Which compound is the limiting reagent? Calculate the number of moles of the excess reagent remaining at the end of the reaction.
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