Hydrogen cyanide, HCN, is prepared from ammonia, air and natural gas \(\left(\mathrm{CH}_{4}\right)\) by the following process. \(2 \mathrm{NH}_{3}(\mathrm{~g})+3 \mathrm{O}_{2}(\mathrm{~g})+2 \mathrm{CH}_{4}(\mathrm{~g}) \stackrel{\mathrm{Pt}}{\longrightarrow}\) \(2 \mathrm{HCN}(\mathrm{g})+6 \mathrm{H}_{2} \mathrm{O}(\mathrm{g})\) If a reaction vessel contains \(11.5 \mathrm{~g} \mathrm{NH}_{3}\), \(10.0 \mathrm{~g} \mathrm{O}_{2}\), and \(10.5 \mathrm{~g} \mathrm{CH}_{4}\), what is the maximum mass, in grams, of hydrogen cyanide that could be made, assuming the reaction goes to completion? (a) \(18.26 \mathrm{~g}\) (b) \(5.625 \mathrm{~g}\) (c) \(17.72 \mathrm{~g}\) (d) \(16.875 \mathrm{~g}\)

Step by step solution

01

Calculate moles of reactants

First, we need to find the molar mass of NH3, O2, and CH4: Molar mass of NH3 = 14.01 (N) + 3 * 1.01 (H) = 17.04 g/mol Molar mass of O2 = 2 * 16.00 (O) = 32.00 g/mol Molar mass of CH4 = 12.01 (C) + 4 * 1.01 (H) = 16.05 g/mol Now, we will determine the number of moles of each reactant present initially: moles of NH3 = 11.5 g / 17.04 g/mol = 0.6747 mol moles of O2 = 10.0 g / 32.00 g/mol = 0.3125 mol moles of CH4 = 10.5 g / 16.05 g/mol = 0.6536 mol

02

Calculate moles of HCN produced by each reactant

Using the stoichiometry of the balanced equation, we will determine the moles of HCN produced by each reactant: For every 2 moles of NH3, 2 moles of HCN are produced. Therefore: moles of HCN (from NH3) = 0.6747 mol * (2 HCN/2 NH3) = 0.6747 mol For every 3 moles of O2, 2 moles of HCN are produced. Therefore: moles of HCN (from O2) = 0.3125 mol * (2 HCN/3 O2) = 0.2083 mol. For every 2 moles of CH4, 2 moles of HCN are produced. Therefore: moles of HCN (from CH4) = 0.6536 mol * (2 HCN/2 CH4) = 0.6536 mol

03

Find the limiting reactant(s) and maximum mass of HCN

We will now determine which reactant yields the least amount of HCN and use that to calculate the maximum mass of HCN that can be produced: The reactant that yields the least HCN is O2, with 0.2083 mol of HCN. Therefore, O2 is the limiting reactant. Now that we know the moles of HCN that can be produced, we can find the maximum mass of HCN: Molar mass of HCN = 12.01 (C) + 14.01 (N) + 1.01 (H) = 27.03 g/mol Maximum mass of HCN = 0.2083 mol * 27.03 g/mol = 5.625 g The maximum mass of hydrogen cyanide that could be made, assuming the reaction goes to completion, is 5.625 g. The correct answer is (b) 5.625 g.

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