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. Mg, 0.16 g
• B. O_2, 0.16 g
• C. Mg, 0.44 g
• D. O₂, 0.28 g

Hint:

The equation is \(2Mg + O_2 → 2MgO\)

Answer 1

The Correct Option is A

To determine which reactant is left in excess and how much is left, we need to analyze the reaction between magnesium (Mg) and oxygen (O₂). The balanced chemical equation for the combustion of magnesium is:

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

From the equation, 2 moles of magnesium react with 1 mole of oxygen gas to produce 2 moles of magnesium oxide.

1. Calculate the number of moles of Mg and O₂:
   • Molar mass of Mg = 24 g/mol
   • Moles of Mg = \frac{{1.0 \text{ g}}}{{24 \text{ g/mol}}} = 0.0417 \text{ mol}
   • Molar mass of O₂ = 32 g/mol
   • Moles of O₂ = \frac{{0.56 \text{ g}}}{{32 \text{ g/mol}}} = 0.0175 \text{ mol}
2. Determine the limiting reactant:

   According to the stoichiometry of the reaction, 2 moles of Mg react with 1 mole of O₂. Therefore, 0.0417 moles of Mg would require \frac{0.0417}{2} = 0.02085 moles of O₂. However, we only have 0.0175 moles of O₂.

   This shows that O₂ is the limiting reactant as there is not enough of it to react with all the Mg present.

3. Calculate the excess amount of Mg:

   Since O₂ is the limiting reactant, it will be completely used up. Using 0.0175 moles of O₂, we can react with 2 \times 0.0175 = 0.035 moles of Mg.

   Excess moles of Mg = 0.0417 - 0.035 = 0.0067 moles

   Mass of excess Mg = 0.0067 \times 24 = 0.16 \text{ g}

Therefore, the reactant left in excess is magnesium (Mg), and the mass of excess magnesium is 0.16 g. Thus, the correct answer is Mg, 0.16 g.