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  • Calculating Theoretical Yields in Chemical Reactions

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    Imagine you’re decorating cupcakes with sprinkles. Each cupcake requires a handful of sprinkles, so you have 12 cupcakes but two large tubs of sprinkles. You have more sprinkles than cupcakes, yet you’ll only finish decorating 12 cupcakes because that’s the limiting factor. In chemistry, the component that limits product formation is called the limiting reagent. Once you identify it, you can compute the theoretical yield—the maximum amount of product you could obtain from the starting materials.

    Identifying the Limiting Reagent

    Consider the reaction that forms ammonia from hydrogen and nitrogen:

    \(\mathrm{H_2+N_2\rightarrow NH_3}\)

    This equation is unbalanced. The balanced form is:

    \(\mathrm{3H_2+N_2\rightarrow 2NH_3}\)

    From the balanced equation we see that 3 moles of hydrogen produce 2 moles of ammonia, and 1 mole of nitrogen also produces 2 moles of ammonia.

    Suppose you start with 4.5 g of hydrogen and 24 g of nitrogen. To determine the limiting reagent, first convert masses to moles using the molar masses (H₂ = 2.02 g mol⁻¹, N₂ = 28.02 g mol⁻¹):

    \(\mathrm{4.5\,g\,H_2\left(\dfrac{1\,mol\,H_2}{2.02\,g\,H_2}\right)=2.23\,mol\,H_2}\)

    \(\mathrm{24\,g\,N_2\left(\dfrac{1\,mol\,N_2}{28.02\,g\,N_2}\right)=0.86\,mol\,N_2}\)

    Using the stoichiometric ratio, find how much nitrogen would be required to consume all 2.23 mol of hydrogen:

    \(\mathrm{2.23\,mol\,H_2\left(\dfrac{1\,mol\,N_2}{3\,mol\,H_2}\right)=0.74\,mol\,N_2}\)

    Similarly, calculate the hydrogen needed for all 0.86 mol of nitrogen:

    \(\mathrm{0.86\,mol\,N_2\left(\dfrac{3\,mol\,H_2}{1\,mol\,N_2}\right)=2.58\,mol\,H_2}\)

    Because you only have 2.23 mol of hydrogen—short of the 2.58 mol required to react with the available nitrogen—hydrogen is the limiting reagent. Once the hydrogen is exhausted, no further ammonia can form, and any remaining nitrogen is left unused, just as excess sprinkles remain after all cupcakes are decorated.

    Calculating the Theoretical Yield

    With hydrogen identified as the limiting reagent, compute the maximum ammonia that can be produced:

    \(\mathrm{2.23\,mol\,H_2\left(\dfrac{2\,mol\,NH_3}{3\,mol\,H_2}\right)=1.49\,mol\,NH_3}\)

    This value, 1.49 mol of ammonia, is the theoretical yield—the highest amount attainable if every molecule of hydrogen reacted perfectly.

    In practice, chemical reactions rarely proceed with 100 % efficiency. Side reactions may consume some of the starting materials, producing unintended products. Thus, the actual yield is typically lower than the theoretical yield, a concept analogous to a sibling stealing a cupcake during decoration.

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