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 <em>limiting reactant</em>. 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 <em>N₂ is the limiting reactant</em>. 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
