Magnesium nitride reacts vigorously with water to form ammonia and magnesium hydroxide, according to the following reaction: blank Mg3N2 plus blank H2O react to produce blank NH3 plus blank Mg(OH)2. Which of the following numbers, in the right order, are the stoichiometric coefficients missing from the reaction equation? (A) One, three, one, two; (B) one, three, two, two; (C) two, six, four, three; (D) two, three, four, three; or (E) one, six, two, three.
Essentially, what we need to do is to balance this equation. Let’s start by rewriting it. We have magnesium nitride, or Mg3N2, reacting with water, or H2O, to produce ammonia, NH3, and magnesium hydroxide, Mg(OH)2. Let’s start by identifying all of the elements present. We have Mg, magnesium; N, nitrogen; H, hydrogen; and O, oxygen. We now need to tally up how many equivalents there are of each atom in the reactants on the left side of the equation and in the products on the right side of the equation.
Let’s start with Mg, magnesium. There’s only one occurrence of magnesium on the reactant side, and it has a subscript of three. So we can do a tally of three next to magnesium in the reactants column. On the right side of the equation, there’s only one occurrence of magnesium and it has no subscript. Thus, there’s only one equivalent.
If we do the same for nitrogen, then we tally two for the reactants and one for the products. If we repeat the process for hydrogen, we tally two for the reactants and then three plus a further two for the product side. Thus, we have a tally of five for hydrogen on the product side. If we do the same again for oxygen, we have one tally on the left and a tally of two on the right.
To balance the reaction equation, we need to make sure that the tallies on each side of the table are equal. Let’s start by looking at magnesium. We have three equivalents for the reactants but only one for the products. If instead we had three moles of magnesium hydroxide, then we’d have three equivalents of magnesium on the product side. Therefore, the tallies would be equal. But in doing this, we also increase the amount of oxygen and hydrogen.
We can count three equivalents of oxygen. But as each equivalent of magnesium hydroxide contains two equivalents of hydroxide ions, we actually need a tally of six for oxygen atoms. The same is true for the number of hydrogen atoms. As there are three equivalents of hydrogen in ammonia, it brings the tally up to a total of nine for hydrogen.
Now, let’s balance the number of nitrogen atoms. We have two equivalents on the reactant side and one on the product side from ammonia. If we have two moles of ammonia, then we’ll have two equivalents of nitrogen on the product side. This is now balanced with the tally of nitrogen on the reactant side. But in doing this, we increase the number of hydrogen atoms. The number is increased by three, bringing the total tally of hydrogen atoms on the product side up to 12.
Let’s now attempt to balance the number of hydrogen atoms. There’s a tally of 12 on the product side but only two on the reactant side. We need to bring this number up to 12. If we add another mole of H2O to the reactant side, we increase the tally of hydrogen by two and oxygen by one. But this still only gives us a tally of four for hydrogen on the reactant side when we need 12. So if we repeat this process of adding equivalents of water to the reactant side, we will eventually have a tally of 12 hydrogen atoms on the reactant side. A total of six moles of water on the reactant side has allowed us to balance the number of hydrogen atoms. In balancing magnesium, nitrogen, and hydrogen, we have inadvertently balanced the number of oxygen atoms.
Now that the number of equivalents of each atom is balanced on both sides of the equation, we can work out the stoichiometric coefficients for the equation. The coefficient for magnesium nitride is one, the coefficient for water is six, the coefficient for ammonia is two, and the coefficient for magnesium hydroxide is three. We have calculated the coefficients to be one, six, two, and three. This matches the values found in option (E).
So the answer to the question “Which of the following numbers, in the right order, are the stoichiometric coefficients missing from the reaction equation?” is (E) one, six, two, three.