Question

Pure magnesium metal is often found as ribbons and can easily burn in the presence of oxygen. When 3.50 g3.50 g of magnesium ribbon burns with 8.91 g8.91 g of oxygen, a bright, white light and a white, powdery product are formed.

1. Enter the balanced chemical equation for this reaction. Be sure to include all physical states.

2. What is the limiting reactant?

3. If the percent yield for the reaction is 79.9%,79.9%, how many grams of product were formed?

4. How many grams of the excess reactant remain?

Answer 1

1.

2Mg(s) + O₂(g) MgO(s)

2 moles of solid Mg reacts with 1 mole of gaseous O₂ to form 2 moles of solid MgO.

2.

Mass of Mg = 3.50 g

Molar mass of Mg = 24.3 g/mol

Moles of Mg = mass of Mg/molar mass of Mg

= 3.50 g/24.3 g/mol

= 0.144 mol

Mass of O₂ = 8.91 g

Molar mass of O₂ = 32.0 g/mol

Moles of O₂ = mass of O₂/molar mass of O₂

= 8.91 g/32.0 g/mol

= 0.278 mol

Required mole ratio of Mg : O₂ for the complete reaction = 2 : 1

Available mole ratio of Mg : O₂ = 0.144 : 0.278

= 1.0 : 1.9

Thus, Mg is in short supply. Hence, Mg is the limiting reagent for the reaction.

3.

Theoretically, 2 mol of Mg gives 2 mol of MgO

Therefore, 0.144 mol of Mg gives = 2 mol x 0.144 mol/2 mol = 0.144 mol of MgO

Molar mass of MgO = 40.3 g/mol

Hence, the theoretical yield of MgO = 0.144 mol x 40.3 g/mol = 5.80 g

Given, percent yield = 79.9 %

Hence, the actual yield = theoretical yield x % yield = 5.80 g x 79.9/100 = 4.63 g

Hence, the amount of products were formed = 4.63 g

4.

Moles of product formed = mass of product/molar mass of product = 4.63 g/40.3 g/mol = 0.115 mol

Now, 2 moles of MgO were formed from 1 mol of O₂

Therefore, 0.115 mol of MgO were formed from = 1 mol x 0.115 mol/2 mol = 0.0575 mol of O₂

Molar mass of O₂ = 32.0 g/mol

Therefore, the mass O₂ used in the reaction = 0.0575 mol x 32.0 g/mol = 1.84 g

Hence, the mass of excess O₂ remained = (8.91 - 1.84) g = 7.07 g

2 moles of MgO is formed from 2 mol of Mg

Therefore, 0.115 mol of MgO is formed from = 2 mol x 0.115 mol/2 mol = 0.115 mol of Mg

Molar mass of Mg = 24.3 g/mol

Therefore, the mass Mg used in the reaction = 0.115 mol x 24.3 g/mol = 2.79 g

Hence, the mass of excess Mg remained = (3.50 - 2.79) g = 0.705 g

Therefore, the total mass of reactant remained = (7.07 + 0.705) g = 7.78 g