Ammonia is produced by the following reaction. 3H2 (g) + N2 (g) - - - > 2NH3 (g) When 7.00 g of hydrogen react with 70.0 g of nitrogen, hydrogen is considered the limiting reactant because

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Chemistry

Thor

23 October, 07:09

Ammonia is produced by the following reaction. 3H2 (g) + N2 (g) - - - > 2NH3 (g) When 7.00 g of hydrogen react with 70.0 g of nitrogen, hydrogen is considered the limiting reactant because

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Savannah Park · Answer 1

Savannah Park 23 October, 07:17

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The correct answer is A

Giovani Underwood · Answer 2

To determine the limiting reactant, we first calculate the number of moles of the reactant.

7.00 g H2 (1 mol / 2.02 g) = 3.47 mol H2

70.00 g N2 (1 mol / 28.02 g) = 2.50 mol N2

From the balanced reaction, we see that there is a 3 is to 1 ratio of the reactants. So, for every 3 moles of H2 we need 1 mole of N2. Given the amount of reactant, for 2.50 moles of N2 we need 7.5 moles of H2 which obviously we have a lesser amount. Hydrogen is the limiting reactant.