To find the theoretical yield of NH3, we need to calculate the limiting reactant in the reaction between H2 and N2.
First, we need to determine the moles of each reactant:
- Moles of H2: 1.52 g / 2.02 g/mol = 0.753 moles
- Moles of N2: 10.1 g / 28.02 g/mol = 0.360 moles
Next, we need to determine the mole ratio between H2 and N2 in the balanced chemical equation:
H2 + N2 -> 2NH3
The ratio is 1:1, which means that H2 is the limiting reactant.
Now, we need to calculate the theoretical yield of NH3 using the moles of H2:
- Moles of NH3 produced = Moles of H2 used = 0.753 moles
- Mass of NH3 produced = Moles of NH3 produced x molar mass of NH3
= 0.753 moles x 17.03 g/mol = 12.81 g
Therefore, the theoretical yield of NH3 in grams is 12.81 g.
What is the theoretical yield in grams for this reaction under the given conditions?
1.52 g H2 is allowed to react with 10.1 g N2, producing 1.26 g NH3.
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