SIDE REACTIONS during the burning of Magnesium 1. Magnesium can react with nitro

Question

SIDE REACTIONS during the burning of Magnesium 1. Magnesium can react with nitrogen gas in air to form magnesium nitride. Write a balanced chemical equation Identify the reaction type- . 2. In an experiment 33.5 g Mg were allowed to react with 28.8 g nitrogen gas, calculate the mass of product 3. What is the theoretical yield of the product Determine the limiting reactant Determine the percent yield of the reaction, if the actual yield is 23.56 g of product. 4. and excess reacant 5. 6 HONORS CHEM: Determine the left-over or unreacted mass of the excess reactant HINT: find the mass of the excess that was needed for reaction by converting grams limiting to grams excess, then subtract that from the original grams of excess reactant

Explanation / Answer

The balanced equation is

3 Mg + N2 ------> Mg3N2

Magnesium metal combines with nitrogen non-metal to form magnesium nitride as product. This is a combination type of reaction.

Number of moles of Mg = 33.5 g / 24.305 g/mol = 1.38 mole

number of moles of N2 = 28.8 g / 28.013 g/mol = 1.03 mol

From the balanced equation we can say that

3 mole of Mg reacts with 1 mole of N2 so

1.38 mole of Mg will react with

= 1.38 mole of Mg *(1 mole of N2 / 3 mole of Mg)

= 0.46 mole of N2

But we have 1.03 mole of N2 which is in excess

so Mg is the limiting reactant

3 mole of Mg produces 1 mole of Mg3N2 so

1.38 mole of Mg will produce

= 1.38 mole of Mg *(1 mole of Mg3N2 / 3 mole of Mg)

= 0.46 mole of Mg3N2

mass of 1 mole of Mg3N2 = 100.9494 g

so the mass of 0.46 mole of Mg3N2 = 46.4 g

Therefore, the mass of product formed would be 46.4 g

3) Theoretical yield of product is 46.4 g

4) limiting reactant is Mg and excess reactant is N2

5) percent yield = (actual yield / theoretical yield)*100

percent yield = (23.56 / 46.4)*100 = 50.8

Therefore, percent yield = 50.8

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