If 28.0g nitrogen gas is reacted with 8.0g of hydrogen gas to form ammonia, the limiting reactant among the two will be:
Correct answer: A. N2
- A. N2
- B. H2
- C. both (a) & (b)
- D. None of these
Explanation
The balanced chemical equation for the production of ammonia is: N2 + 3H2 → 2NH3. From this equation, 1 mole of nitrogen reacts with 3 moles of hydrogen. Calculate the moles of each reactant: 28.0g of N2 is 1 mole (since the molar mass of N2 is 28.0 g/mol), and 8.0g of H2 is approximately 4 moles (since the molar mass of H2 is 2.0 g/mol). Given this, the reaction requires 3 moles of H2 for every mole of N2. Hence, the 4 moles of H2 are in excess, making N2 the limiting reactant. Option A is correct. Options B, C, and D are incorrect because hydrogen is in excess, and only one reactant can be the limiting reactant.
This question appeared on the ETEA MDCAT 2006 paper, which you can sit online with every answer explained.
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About Limiting and Excess Reactants
Limiting and excess reactants are identified from the balanced chemical equation and the available amounts of each substance. The limiting reactant determines the maximum product formed, while the excess reactant remains after completion, so questions require mole ratios, theoretical yield and sometimes percentage yield calculations.
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