Chemistry: For the following reaction, 15.7 grams of magnesium nitride are allowed to react with 13.2 grams of water.

Question

Magnesium nitride (s) + water (l) —> magnesium hydroxide (aq) + ammonia (aq)

What is the maximum amount of magnesium hydroxide that can be formed?

Mass=

What is the formula for the limiting reactant?

What amount of the excess reactant remains after the reaction is complete?

Mass=

Answer 1

First we convert names to chemical formulas, balance the equation, and indicate given quantities

Mg₃N₂ + 6 H₂O -—> 3 Mg(OH)₂ + 2 NH₃

15.7 g 13.2 g

Next we convert these quantities to moles using molar masses

Mg₃N₂ (100.9494 g/mol) and H₂O (18 g/mol)

Dividing an amount in g by molar mass converts it to an amount in moles:

for Mg₃N₂ (15.7 g) / (100.9494 g/mol) = 0.1555 mol

for H₂O (13.2 g) / (18 g/mol) = 0.7333 mol

The stoichiometry tells us we need 6 × 0.1555 mol = 0.933 mol of H₂O to react with all of the Mg₃N₂. Since we have less H₂O that this, H₂O is the limiting reagent.

0.7333 mol H₂O will convert ⅙(0.7333) = 0.1222 mol Mg₃N₂ into ½(0.7333) = 0.3667 mol Mg(OH)₂

This leaves 0.1555 – ⅙(0.7333) = 0.0333 mol Mg₃N₂ unreacted.

Finally, we multiply these amounts in moles by molar mass to convert them to amounts in g. We need one more molar mass Mg(OH)₂ (58.3197 g/mol)

for the amount of Mg(OH)₂ formed: 0.3667 mol × 58.3197 g/mol = 21.38 g

for the amount of Mg₃N₂ left unreacted: 0.0333 mol × 100.9494 g/mol = 3.36 g

Answers

The maximum amount of magnesium hydroxide that can be formed is 21.38 g

The formula for the limiting reactant is H₂O

The amount of the excess reactant (Mg₃N₂) that remains after the reaction is complete is 3.36 g

Answer 2

The balanced chemical reaction with magnesium nitride and water to produce magnesium hydroxide and ammonia is: Mg₃N₂(s) + 6H₂O(l) → 3Mg(OH)₂(aq) + 2NH₃(aq)

Using the periodic table to find the molecular weight of each element in this chemical reaction.

Mg: 24.305 g/mol

N: 14.007 g/mol

H: 1.008 g/mol

O: 15.999 g/mol

Now determine the molecular weight of the reactants and products. For example, 1 mole Mg₃N₂ consists of 3 moles of Mg and 2 moles of N. The weight of 1 mol of Mg₃N₂ is therefore:

(3 mol Mg)(24.305 g/mol Mg) + (2 mol N)(14.007 g/mol N) = 100.929 g

The other molar weights of interest are as follows

H₂O: (2 mol H)(1.008 g/mol) + (1 mol O)(15.999 g/mol) = 18.015 g

Mg(OH)₂: (1 mol Mg)(24.305 g/mol) + (2 mol OH)(17.007 g/mol) = 58.319 g

NH₃: (1 mol N)(14.007 g/mol) + (3 mol H)(1.008 g/mol) = 17.031 g

We need to determine whether Mg₃N₂(s) or H₂O(l) is the limiting reactant. We are told that 15.7 g Mg₃N₂(s) was allowed to react with 13.2 g H₂O. The balanced chemical reaction tells us that 1 mole of Mg₃N₂(s) will react with 6 moles of H₂O.

Let's start solving this problem by converting the amount of Mg₃N₂(s) used from grams to moles:

(15.7 g Mg₃N₂(s)) x (1 mol Mg₃N₂(s) ÷ 100.929 g) = 0.156 mol Mg₃N₂(s)

The number of moles of water that would react 0.156 moles of Mg₃N₂(s) is

(0.156 mol Mg₃N₂(s)) x (6 mol H₂O ÷ 1 mol Mg₃N₂(s)) = 0.933 mol H₂O

Now, let's convert the amount of water from moles to grams.

(0.933 mol H₂O) x (18.015 g/mol) = 16.8 g H₂O

Since this is greater than the 13.2 g H₂O that was actually used, we can determine that H₂O is the limiting reactant.

To determine the amount of excess Mg₃N₂(s) leftover after the reaction is complete, we will calculate the amount of Mg₃N₂(s) that reacted with water and subtract this from the amount of Mg₃N₂(s) that was present at the start of the reaction (15.7 g).

The amount of Mg₃N₂(s) that reacted with water is calculated as follows:

(13.2 g H₂O) x (1 mol H₂O ÷ 18.015 g H₂O) x (1 mol Mg₃N₂(s) ÷ 6 mol H₂O) x (100.929 g Mg₃N₂(s) ÷ 1 mol Mg₃N₂(s)) = 12.3 g Mg₃N₂(s)

The amount of Mg₃N₂(s) leftover is 15.7 g - 12.3 g = 3.4 g

The amount of Mg(OH)₂ formed can be calculated using the balanced chemical reaction

Mg₃N₂(s) + 6H₂O(l) → 3Mg(OH)₂(aq) + 2NH₃(aq)

You can start with either the amount of Mg₃N₂(s) or water that reacted. Since 13.2 g H₂O was the limiting reactant, the amount formed is:

(13.2 g H₂O) x (1 mol H₂O ÷ 18.015 g H₂O) x (3 mol Mg(OH)₂ ÷ 6 mol H₂O) x (58.319 g Mg(OH)₂ ÷ 1 mol Mg(OH)₂) = 21.4 g Mg(OH)₂