How many moles of NH₃ can be produced by the reaction of 2.00 g of N₂ with 3.00 g H₂?
Reaction: N₂(g) + 3 H₂(g) → 2 NH₃(g)

Question

How many moles of NH₃ can be produced by the reaction of 2.00 g of N₂ with 3.00 g H₂? Reaction: N₂(g)+ 3 H₂(g)→ 2 NH₃(g)

Quizplus · Accepted Answer

The balanced chemical equation is:N2(g) + 3 H2(g) → 2 NH3(g)From the equation, we can see that 1 mole of N2 reacts with 3 moles of H2 to produce 2 moles of NH3.To find the limiting reactant, we need to compare the number of moles of each reactant to the stoichiometric ratio.For N2:2.00 g N2 * (1 mol N2 / 28.01 g N2) = 0.0714 mol N2For H2:3.00 g H2 * (1 mol H2 / 2.02 g H2) = 1.49 mol H2Comparing the number of moles of each reactant to the stoichiometric ratio, we find that N2 is the limiting reactant because it has the smallest number of moles relative to the stoichiometric ratio.Therefore, the number of moles of NH3 that can be produced is:0.0714 mol N2 * (2 mol NH3 / 1 mol N2) = 0.143 mol NH3So, the correct answer is A) 0.143.

How Many Moles of NH3 Can Be Produced by the Reaction

How Many Moles of NH₃ Can Be Produced by the Reaction