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

3H2(g) + N2 --> 2 NH3

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

adding or remove a gaseous reactant or product changes the concentrations. if the concentration of reactant or product is increased, the system will shift away from the side in which concentration was increased (i.e. if the concentration of reactants is increased, the system will shift toward the products. if more products are added, the system will shift to form more reactants). conversely, if the concentration of reactant or product is decreased, the system will shift toward the side in which concentration was decreased (i.e. if reactants are removed, the system will shift to form more reactants. if the concentration of products is decreased, the equilibrium will shift toward the products).

explanation:

your answer would be a. 12.5 kg