Question

1.0 g of magnesium is burnt with 0.56 g O₂ in a closed vessel. Which reactant is left in excess and how much ?
(At. wt. Mg =24; O=16)

• A. O₂, 0.16 g
• B. Mg, 0.44 g
• C. O₂, 0.28 g
• D. Mg, 0.16 g

Answer 1

The Correct Option is D

To determine which reactant is left in excess, we need to find the limiting reactant in the reaction between magnesium (Mg) and oxygen (O₂) to form magnesium oxide (MgO). The balanced chemical equation for the reaction is:

2 \text{Mg} + \text{O}_2 \rightarrow 2 \text{MgO}

Step 1: Calculate the number of moles of each reactant.

For magnesium (Mg):

Given mass of Mg = 1.0 g

Molar mass of Mg = 24 g/mol

Number of moles of Mg = \frac{1.0}{24} = 0.0417 moles

For oxygen (O₂):

Given mass of O₂ = 0.56 g

Molar mass of O₂ = 32 g/mol

Number of moles of O₂ = \frac{0.56}{32} = 0.0175 moles

Step 2: Determine the limiting reactant.

The stoichiometry of the reaction shows that 2 moles of Mg react with 1 mole of O₂.

Therefore, the mole ratio required for Mg to O₂ is 2:1.

From the calculated moles:

Available mole ratio of Mg to O₂ = \frac{0.0417}{0.0175} \approx 2.38

This is greater than 2:1, indicating that Mg is in excess and O₂ is the limiting reactant.

Step 3: Calculate the amount of excess Mg.

Moles of Mg required to react with 0.0175 moles of O₂ = 2 \times 0.0175 = 0.035 moles

Moles of excess Mg = 0.0417 - 0.035 = 0.0067 moles

Mass of excess Mg = 0.0067 \times 24 = 0.16 g

Conclusion: Magnesium (Mg) is left in excess, and the amount of excess Mg is 0.16 g.