Question

What is the result when 12 grams of H2 and 28 grams of N2 react to completion at STP.

Answer 1

Answer:

The correct answer is 89.6 L

Explanation:

We have the following chemical equation and the molar masses for the reaction:

3H₂(g)   +   N₂ -->   2 NH₃

 6 g            28 g      34 g

That means that 3 moles of H₂ (6 g) reacts with 1 mol of N₂ (28 g) and gives 2 moles of NH₃ (34 g). In order to calculate how many liters of NH₃ result from the reaction of 12 grams of H₂ and 28 grams of N₂, we have to first figure out which reactant is the limiting reactant. According to the equation, if 6 grams of H₂ reacts with 28 g of N₂, and we have 12 grams:

6 g H₂------- 28 g N₂

12 g H₂-------- X = 12 g H₂ x 28 g N₂/6 g H₂ = 56 g N₂

We need 56 g of N₂ but we have 28 g of N₂, so N₂ is the limiting reactant. With the limiting reactant we can calculate the moles of product (NH₃) we will obtain:

We have 28 g N₂ -----> 28 g/14 g/mol = 2 moles N₂

1 mol N₂ ----------- 2 moles NH₃

2 mol N₂ --------- X = 2 mol N₂ x 2 moles NH₃/1 mol N₂ = 4 mol NH₃

Finally, we convert the moles of NH₃ to liters:

1 mol gas at STP = 22.4 L

Liters NH₃ obtained = 4 moles NH₃ x 22.4 L/1 mol = 89.6 L