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Chapter 3: Problem 93

When heated, lithium reacts with nitrogen to form lithium nitride: $$6 \mathrm{Li}(s)+\mathrm{N}_{2}(g) \longrightarrow 2 \mathrm{Li}_{3} \mathrm{~N}(s)$$What is the theoretical yield of \(\mathrm{Li}_{3} \mathrm{~N}\) in grams when \(12.3 \mathrm{~g}\) of \(\mathrm{Li}\) are heated with \(33.6 \mathrm{~g}\) of \(\mathrm{N}_{2} ?\) If the actual yield of \(\mathrm{Li}_{3} \mathrm{~N}\) is \(5.89 \mathrm{~g},\) what is the percent yield of the reaction?

Short Answer

Expert verified
The theoretical yield of \(Li_3N\) is approximately 83.6 g and the percent yield of the reaction is approximately 7.05%.

Step by step solution

01

Determine the limiting reagent

To determine the limiting reagent, we have to compare the mole-to-mole ratio of the reactants to the coefficients in the balanced chemical equation. First, we need to convert the given mass of each reactant into moles. \n\nThe molar mass of \(Li\) is approximately \(6.94 g/mol\) and that of \(N_2\) is approximately \(28.01 g/mol\). So, the number of moles of \(Li\) is \(12.3 g / 6.94 g/mol \approx 1.77 mol\) and for \(N_2\) it's \(33.6 g / 28.01 g/mol \approx 1.20 mol\). \n\nThe stoichiometric ratio from the balanced chemical equation for \(Li\) and \(N_2\) is \(6:1\). Therefore, in terms of \(N_2\), we have enough \(Li\) because \(1.77 mol > 1.20 mol\). So, \(N_2\) is the limiting reagent.
02

Calculate the theoretical yield

Once we know the limiting reagent, we can estimate the maximum amount of product that could be formed, known as the theoretical yield. The stoichiometric ratio between \(N_2\) and \(Li_3N\) is \(1:2\), hence for each mole of \(N_2\), we will have two moles of \(Li_3N\).\n\nGiven that the number of moles of \(N_2\) is roughly \(1.20 mol\), the number of moles of \(Li_3N\) is \(2 × 1.20 mol = 2.40 mol\). To convert this to grams, we must multiply by the molar mass of \(Li_3N\) which is roughly \(34.83 g/mol\). \n\nWe then get the theoretical yield of \(Li_3N\) to be \(2.40 mol × 34.83 g/mol = 83.6 g\).
03

Calculate the percent yield

The percent yield is the ratio of the actual yield to the theoretical yield, multiplied by 100%. Given that the actual yield of \(Li_3N\) is \(5.89 g\), the percent yield would be \((5.89 g / 83.6 g) × 100% \approx 7.05%\).

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Most popular questions from this chapter

A compound containing only \(\mathrm{C}, \mathrm{H},\) and \(\mathrm{Cl}\) was examined in a mass spectrometer. The highest mass peak seen corresponds to an ion mass of 52 amu. The most abundant mass peak seen corresponds to an ion mass of 50 amu and is about three times as intense as the peak at 52 amu. Deduce a reasonable molecular formula for the compound and explain the positions and intensities of the mass peaks mentioned.

A compound made up of \(\mathrm{C}, \mathrm{H},\) and \(\mathrm{Cl}\) contains 55.0 percent \(\mathrm{Cl}\) by mass. If \(9.00 \mathrm{~g}\) of the compound contain \(4.19 \times 10^{23} \mathrm{H}\) atoms, what is the empirical formula of the compound?

(a) For molecules having small molecular masses, mass spectrometry can be used to identify their formulas. To illustrate this point, identify the molecule that most likely accounts for the observation of a peak in a mass spectrum at 16 amu, 17 amu, 18 amu, and 64 amu. (b) Note that there are (among others) two likely molecules that would give rise to a peak at 44 amu, namely, \(\mathrm{C}_{3} \mathrm{H}_{8}\) and \(\mathrm{CO}_{2} .\) In such cases, a chemist might try to look for other peaks generated when some of the molecules break apart in the spectrometer. For example, if a chemist sees a peak at 44 amu and also one at 15 amu, which molecule is producing the 44 -amu peak? Why? (c) Using the following precise atomic masses \(-\mathrm{H}\) ( 1.00797 amu), \({ }^{12} \mathrm{C}(12.00000 \mathrm{amu}),\) and \({ }^{16} \mathrm{O}(15.99491 \mathrm{amu})-\) how precisely must the masses of \(\mathrm{C}_{3} \mathrm{H}_{8}\) and \(\mathrm{CO}_{2}\), be measured to distinguish between them?

Lysine, an essential amino acid in the human body, contains \(\mathrm{C}, \mathrm{H}, \mathrm{O},\) and \(\mathrm{N} .\) In one experiment, the complete combustion of \(2.175 \mathrm{~g}\) of lysine gave \(3.94 \mathrm{~g}\) \(\mathrm{CO}_{2}\) and \(1.89 \mathrm{~g} \mathrm{H}_{2} \mathrm{O} .\) In a separate experiment, \(1.873 \mathrm{~g}\) of lysine gave \(0.436 \mathrm{~g} \mathrm{NH}_{3}\). (a) Calculate the empirical formula of lysine. (b) The approximate molar mass of lysine is \(150 \mathrm{~g}\). What is the molecular formula of the compound?

A sample containing \(\mathrm{NaCl}, \mathrm{Na}_{2} \mathrm{SO}_{4},\) and \(\mathrm{NaNO}_{3}\) gives the following elemental analysis: Na: 32.08 percent; O: 36.01 percent; Cl: 19.51 percent. Calculate the mass percent of each compound in the sample.
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