Consider the reaction 2 NO(g) + 5 H2(g) → 2 NH3(g) + 2 H20(g) The reaction mixture initially contains 6 moles of NO and 10 moles of H2 contained in a 20.0- liter gas cylinder with a movable piston. At the end of the reaction, the volume has decreased to 15.7 L. Without doing any calculations, which method is the best for determining the concentration of NO in the mixture after the reactants have reacted as completely as possible? Use stoichiometry: 10 moles of H2 consumes 6 moles of NO, so the final concentration of NO = 0 M Use the dilution formula: (6 mol NO)(20.0 L) = (x mol NO)(15.7 L) Use the dilution formula: (2 mol NO)(20.0 L) = (x mol NO)(15.7 L) Use stoichiometry: 10 moles of H2 consumes 4 moles of NO, so the final concentration of NO = moles - 4 moles NO)/(15.7 L) (6

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Consider the reaction
2 NO(g) + 5 H2(g)
→ 2 NH3(g) + 2 H20(g)
The reaction mixture initially contains 6 moles of NO and 10 moles of H2 contained in a 20.0-
liter gas cylinder with a movable piston. At the end of the reaction, the volume has decreased
to 15.7 L. Without doing any calculations, which method is the best for determining the
concentration of NO in the mixture after the reactants have reacted as completely as
possible?
Use stoichiometry: 10 moles of H2 consumes 6 moles of NO, so the final concentration of NO = 0 M
Use the dilution formula: (6 mol NO)(20.0 L) = (x mol NO)(15.7 L)
Use the dilution formula: (2 mol NO)(20.0 L) = (x mol NO)(15.7 L)
Use stoichiometry: 10 moles of H2 consumes 4 moles of NO, so the final concentration of NO = (6
moles - 4 moles NO)/(15.7 L)
Transcribed Image Text:Consider the reaction 2 NO(g) + 5 H2(g) → 2 NH3(g) + 2 H20(g) The reaction mixture initially contains 6 moles of NO and 10 moles of H2 contained in a 20.0- liter gas cylinder with a movable piston. At the end of the reaction, the volume has decreased to 15.7 L. Without doing any calculations, which method is the best for determining the concentration of NO in the mixture after the reactants have reacted as completely as possible? Use stoichiometry: 10 moles of H2 consumes 6 moles of NO, so the final concentration of NO = 0 M Use the dilution formula: (6 mol NO)(20.0 L) = (x mol NO)(15.7 L) Use the dilution formula: (2 mol NO)(20.0 L) = (x mol NO)(15.7 L) Use stoichiometry: 10 moles of H2 consumes 4 moles of NO, so the final concentration of NO = (6 moles - 4 moles NO)/(15.7 L)
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