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Chapter 3: Problem 121

Hydrogen cyanide is produced industrially from the reaction of gaseous ammonia, oxygen, and methane: \(2 \mathrm{NH}_{3}(g)+3 \mathrm{O}_{2}(g)+2 \mathrm{CH}_{4}(g) \longrightarrow 2 \mathrm{HCN}(g)+6 \mathrm{H}_{2} \mathrm{O}(g)\) If \(5.00 \times 10^{3} \mathrm{~kg}\) each of \(\mathrm{NH}_{3}, \mathrm{O}_{2}\), and \(\mathrm{CH}_{4}\) are reacted, what mass of \(\mathrm{HCN}\) and of \(\mathrm{H}_{2} \mathrm{O}\) will be produced, assuming \(100 \%\) yield?

Short Answer

Expert verified
Assuming 100% yield, 2813 kg of HCN and 5632 kg of H2O will be produced.

Step by step solution

01

Calculate the moles of each reactant

First, we need to find the number of moles of each reactant. To do this, we can use the molar mass of each reactant and the given mass: NH3: \(M_{NH3} = 14.01 + 3(1.008) = 17.03~\text{g/mol}\) \\ O2: \(M_{O2} = 2(16.00) = 32.00~\text{g/mol}\) \\ CH4: \(M_{CH4} = 12.01 + 4(1.008) = 16.04~\text{g/mol}\) And then divide the mass by the molar mass: moles of NH3: \(\frac{5.00 \times 10^3~\text{kg}}{17.03~\text{g/mol}} = 293.6 \times 10^3~\text{moles}\) \\ moles of O2: \(\frac{5.00 \times 10^3~\text{kg}}{32.00~\text{g/mol}} = 156.25 \times 10^3~\text{moles}\) \\ moles of CH4: \(\frac{5.00 \times 10^3~\text{kg}}{16.04~\text{g/mol}} = 311.82 \times 10^3~\text{moles}\)
02

Determine the limiting reactant

Next, we need to determine which reactant is the limiting reactant, as it will determine the formation of HCN and H2O. By comparing their mole ratios, we can find the limiting reactant: NH3/O2: \(\frac{293.6 \times 10^3~\text{moles}}{2} / \frac{156.25 \times 10^3~\text{moles}}{3} = 1.25\) \\ O2/CH4: \(\frac{156.25 \times 10^3~\text{moles}}{3} / \frac{311.82 \times 10^3~\text{moles}}{2} = 0.667\) Since the NH3/O2 ratio is greater than the stoichiometric ratio (2:3) and the O2/CH4 ratio is less than the stoichiometric ratio (3:2), O2 is the limiting reactant.
03

Calculate the moles of each product

Now that we know O2 is the limiting reactant, we can calculate the moles of each product formed based on the balanced chemical equation and the stoichiometric ratios: moles of HCN: \(\frac{2}{3} \times (156.25 \times 10^3~\text{moles}) = 104.17 \times 10^3~\text{moles}\) \\ moles of H2O: \(\frac{6}{3} \times (156.25 \times 10^3~\text{moles}) = 312.5 \times 10^3~\text{moles}\)
04

Calculate the mass of each product

Lastly, we need to convert the number of moles of each product back to mass. To do this, multiply the moles by the molar mass of each product: HCN: \(M_{HCN} = 12.01 + 14.01 + 1.008 = 27.03~\text{g/mol}\) \\ H2O: \(M_{H2O} = 2(1.008) + 16.00 = 18.02~\text{g/mol}\) mass of HCN: \((104.17 \times 10^3~\text{moles}) \times (27.03~\text{g/mol}) = 2813 \times 10^3~\text{g}\) \\ mass of H2O: \((312.5 \times 10^3~\text{moles}) \times (18.02~\text{g/mol}) = 5632 \times 10^3~\text{g}\) So, assuming 100% yield, 2813 kg of HCN and 5632 kg of H2O will be produced.

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Most popular questions from this chapter

In 1987 the first substance to act as a superconductor at a temperature above that of liquid nitrogen \((77 \mathrm{~K})\) was discovered. The approximate formula of this substance is \(\mathrm{YBa}_{2} \mathrm{Cu}_{3} \mathrm{O}_{7}\). Calculate the percent composition by mass of this material.

Arrange the following substances in order of increasing mass percent of carbon. a. caffeine, \(\mathrm{C}_{8} \mathrm{H}_{10} \mathrm{~N}_{4} \mathrm{O}_{2}\) b. sucrose, \(\bar{C}_{12} \mathrm{H}_{22} \mathrm{O}_{11}\) c. ethanol, \(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}\)

A sample of urea contains \(1.121 \mathrm{~g} \mathrm{~N}, 0.161 \mathrm{~g} \mathrm{H}, 0.480 \mathrm{~g} \mathrm{C}\), and \(0.640 \mathrm{~g} \mathrm{O}\). What is the empirical formula of urea?

Aspirin \(\left(\mathrm{C}_{9} \mathrm{H}_{8} \mathrm{O}_{4}\right)\) is synthesized by reacting salicylic acid \(\left(\mathrm{C}_{7} \mathrm{H}_{6} \mathrm{O}_{3}\right)\) with acetic anhydride \(\left(\mathrm{C}_{4} \mathrm{H}_{6} \mathrm{O}_{3}\right)\). The balanced equa- tion is $$ \mathrm{C}_{7} \mathrm{H}_{6} \mathrm{O}_{3}+\mathrm{C}_{4} \mathrm{H}_{6} \mathrm{O}_{3} \longrightarrow \mathrm{C}_{9} \mathrm{H}_{3} \mathrm{O}_{4}+\mathrm{HC}_{2} \mathrm{H}_{3} \mathrm{O}_{2} $$ a. What mass of acetic anhydride is needed to completely consume \(1.00 \times 10^{2} \mathrm{~g}\) salicylic acid? b. What is the maximum mass of aspirin (the theoretical yield) that could be produced in this reaction?

Silver sulfadiazine burn-treating cream creates a barrier against bacterial invasion and releases antimicrobial agents directly into the wound. If \(25.0 \mathrm{~g} \mathrm{Ag}_{2} \mathrm{O}\) is reacted with \(50.0 \mathrm{~g} \mathrm{C}_{10} \mathrm{H}_{10} \mathrm{~N}_{4} \mathrm{SO}_{2}\), what mass of silver sulfadiazine, \(\mathrm{AgC}_{10} \mathrm{H}_{9} \mathrm{~N}_{4} \mathrm{SO}_{2}\), can be produced, assuming \(100 \%\) yield? $$ \mathrm{Ag}_{2} \mathrm{O}(s)+2 \mathrm{C}_{10} \mathrm{H}_{10} \mathrm{~N}_{4} \mathrm{SO}_{2}(s) \longrightarrow 2 \mathrm{AgC}_{10} \mathrm{H}_{9} \mathrm{~N}_{4} \mathrm{SO}_{2}(s)+\mathrm{H}_{2} \mathrm{O}(l) $$
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