Question 3.3.3: Use balanced chemical equations to relate amounts of reactan......

You are asked to balance a chemical equation and determine the amount of reactant consumed and products formed by a given amount of reactant.
You are given an unbalanced equation and the amount of one reactant consumed in the reaction.
Step 1. Write a balanced chemical equation.

Mg_{3}N_{2}(s) + 4 H_{2}SO_{4}(aq) →  3  MgSO_{4}(aq) + (NH_{4})_{2}SO_{4}(aq)

Step 2. Use the coefficients in the balanced equation to create a stoichiometric factor that will convert moles of Mg_{3}N_{2} to moles of H_{2}SO_{4} consumed.

3.5 mol Mg_{3}N_{2}\times \frac{4\text{ mol H}_{2}\text{SO}_{4}}{1\text{ mol Mg}_{3}\text{N}_{2}} = 14 mol H_{2}SO_{4} consumed

Step 3. Use the coefficients in the balanced equation to create a stoichiometric factor that will convert moles of Mg_{3}N_{2} to moles of MgSO_{4} produced.

3.5 mol Mg_{3}N_{2}\times \frac{3\text{ mol MgSO}_{4}}{1\text{ mol Mg}_{3}\text{N}_{2}} = 11 mol MgSO_{4}

Step 4. Use the coefficients in the balanced equation to create a stoichiometric factor that will convert moles of Mg_{3}N_{2} to moles of (NH_{4})_{2}SO_{4} produced.

3.5 mol Mg_{3}N_{2}\times \frac{1\text{ mol (NH}_{4}\text{)}_{2}\text{SO}_{4}}{1 \text{ mol Mg}_{3}\text{ N}_{2}} = 3.5 mol (NH_{4})_{2}SO_{4}

Is your answer reasonable? According to the balanced equation, one mole of Mg_{3}N_{2} reacts with four moles of H_{2}SO_{4}, producing three moles of MgSO_{4} and one mole of (NH_{4})_{2}SO_{4}. Here, more than one mole of Mg_{3}N_{2} reacts, which means more than these amounts of reactants and products are consumed and produced, respectively