Consider the following reaction: 2SO2(g) + O2(g) --> 2SO3(g) If 282.7 mL of SO2

Question

Consider the following reaction: 2SO2(g) + O2(g) --> 2SO3(g)

If 282.7 mL of SO2 is allowed to react with 165.0 mL of O2 (both measured at 317 K and 46.6 mmHg), what is the limiting reactant?

I know the answer which is that SO2 is the limiting reactant but I don't know how to get it.

Also what is the theoretical yield of SO3? How do I go about setting up an equation to solve it?

Explanation / Answer

Use the ideal gas equation, PV = nRT, to solve for the moles of each of the reactants: Moles of SO2: P = 46.6 mmHg = 0.0613 atm V = 282.7 mL = 0.2827 L R = 0.0821 atm*L/mol*K (gas constant) T = 317 K n = ? PV = nRT (0.0613 atm)(0.2827 L) = n(0.0821 atm*L/mol*K)(317 K) n = [(0.0613 atm)(0.2827 L)] / (0.0821 atm*L/mol*K)(317 K) n = 6.66 x 10^-4 moles Moles of O2: P = 46.6 mmHg = 0.0613 atm V = 165.0 mL = 0.165 L R = 0.0821 atm*L/mol*K (gas constant) T = 317 K n = ? PV = nRT (0.0613 atm)(0.165 L) = n(0.0821 atm*L/mol*K)(317 K) n = [(0.0613 atm)(0.165 L)] / (0.0821 atm*L/mol*K)(317 K) n = 3.89 x 10^-4 moles To find out the LR, we must find the moles of SO3 from the moles of each of the reactants: Moles of SO3 from moles of SO2 --> Ratio of SO2:SO3 = 2:2, thefore moles of SO3 = 6.66 x 10^-4 moles Moles of SO3 from moles of O2 --> Ratio of O2:SO3 = 1:2, thefore moles of SO3 = 7.78 x 10^-4 moles Since SO2 produces the fewer amount of moles of SO3, it is the LR. Theoretical yield of SO3 = 6.66 x 10^-4 moles (calculated above) Hope this helps! :-)

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