Metallic magnesium reacts with steam to produce magnesium hydroxide and hydrogen gas. a. If 16.2 g Mg are heated with 12.0 g H2O, what is the limiting reactant? b. How many moles of the excess reactant are left? c. How many grams of each product are formed?

a) Limiting reactant: metallic magnesium b) Excess: 0.076 moles c) Mg (OH) ₂ : 17.2 g d) H₂ (g) : 0.595g

Explanation:

a) What is the limiting reactant

1. Write the balanced molecular equation

Mg (s) + 2H₂O (g) → Mg (OH) ₂ (s) + H₂ (g) ↑

2. Write the mole ratio of the reactants

1 mol Mg : 2 mol H₂O ⇒ 1 : 2

3. Calculate the number of moles of both reactants

number of moles = mass in grams / molar mass

i) Mg:

number of moles = 16.2g / (54.938g/mol) ≈ 0.295 mol

ii) H₂O

number of moles = 12.0g / (18.015g/mol) ≈ 0.666 mol

4. Limiting reactant

Since the theoretical mole ratio is 1 : 2, 0.295 moles of Mg react with 2 * 0.295 = 0.590 moles of H₂O (steam).

Then, since there are 0.666 moles of steam available, all the Mg can react and it is the limiting reactant.

b) How many moles of the excess reactant are left?

The excess reactant is the steam.

As determined above, since the mole ratio is 1 : 2, 0.590 moles of steam will react with all the available metallic magnesium. Thus, the excess is:

0.666mol - 0.590mol = 0.076 moles left of the excess reactant.

c) How many moles of each product are formed?

You must use the number of moles of the limiting reactant: 0.295 moles of Mg.

i) Mg (OH)

Mole ratio: 1 mol Mg : 1 mol Mg (OH) ₂

Then, 0.295 moles of Mg (OH) ₂ are formed.

Convert to grams:

mass = number of moles * molar mass mass = 58.3197g/mol mass = 0.295 mol * 58.3197 g/mol = 17.2 g.

ii) H₂ (g)

Mole ratio: 1 mol Mg : 1 mol H₂ (g)

Then, 0.295 moles of H₂ (g) are formed

mass = number of moles * molar mass mass = 0.295 mol * 2.016g/mol = 0.595g