According to the following table, which mixture can we use to obtain the maximum mass of precipitated Ca₃(Po₄)₂? A) mixture I, B) II, C) III, D) IV, or E, V.
The question gives us a table with data for Ca(NO₃)₂, calcium nitrate, and Na₃PO₄, sodium phosphate. The first thing to do is to write an equation for the reaction between calcium nitrate and sodium phosphate and then balance it. The reaction is as follows. Calcium nitrate reacts with sodium phosphate to give Ca₃(PO₄)₂, which is calcium phosphate, and NaNO₃, sodium nitrate.
The question tells us that calcium phosphate is precipitated. This means that it is formed as an insoluble solid product. If a product is described as a precipitate, this means that the reactants that reacted to produce it must both have been in the aqueous phase. We can assume that sodium nitrate is also in the aqueous phase or soluble in water and that the only insoluble product is calcium phosphate.
Next, we need to balance the equation. When we balance the equation, we get the stochiometric coefficients of three, two, one, and six, respectively. We say the mole ratio of the reaction is three as to two as to one as to six. This means that three moles of calcium nitrate reacts with two moles of sodium phosphate to produce one mole of calcium phosphate and six moles of sodium nitrate. There are several ways to determine the maximum mass of precipitated product. One way is to determine the number of moles of each reactant, then decide which of the reactants is the limiting reagent. It is the limiting reagent which determines the number of moles of product that can form. Then we determine the number of moles of product and, from this, the mass of product.
So let’s begin by determining the number of moles of each reactant. Number of moles is equal to concentration times volume. For the calcium nitrate reactant, we are given a concentration of 0.5 M or 0.5 molar. When the unit for concentration is a capital M, we can call concentration molarity, which really means a concentration in terms of moles per liter, in other words, a unit of moles per liter. Sodium phosphate also has a molarity of 0.5 molar or 0.5 moles per liter. So both reactants have the same concentration.
Because we are not allowed to use calculators here, let’s simplify things by making 0.5 M equal to 𝑥. So for mixture I, the number of moles of calcium nitrate is equal to 𝑐 times 𝑣. Substituting 𝑥 for 𝑐 and three for volume, we get three 𝑥 moles. Note that usually we would have to convert the volume value of three milliliters into a liter value by dividing by 1000. But because we are calling the concentration value 𝑥, we do not need to consider the volume unit. So the number of moles of calcium nitrate in mixture I is three 𝑥.
We can do the same calculation for all five mixtures for each reactant. And we get — we now have the number of moles in terms of 𝑥 for each reagent in each mixture. The next step is to determine which reactant or reagent is limiting. It is the limiting reagent which determines or controls the number of moles of product that can form. One way to determine the limiting reagent is to calculate the number of moles of product that can form from each of the reactants. After this, we will compare the number of moles of product formed from each reactant to see which is the smaller value in each mixture. And the smaller value will indicate to us the limiting reagent for each mixture.
So for mixture I, let’s calculate the number of moles of calcium phosphate product that can form from calcium nitrate reactant. We know that the calcium nitrate reactant produces a number of moles of calcium phosphate product three times smaller. So we divide the number of moles of calcium nitrate by three to get the number of moles of precipitate product. In mixture I, the number of moles or precipitate product is one 𝑥.
Let’s do the same for all the mixtures. Let’s now determine the number of moles of calcium phosphate product from the number of moles of sodium phosphate reactant. According to the mole ratio, we know that the number of moles of calcium phosphate product will be half that of sodium phosphate reactant. So we can divide the number of moles of sodium phosphate by two. And we get the number of moles of calcium phosphate that can form from sodium phosphate in terms of 𝑥.
We are still determining which is the limiting reactant. Now, we need to compare the moles of product that can form from each reactant. In mixture I, 𝑥 is smaller than four 𝑥. And it is the smaller value which indicates the limiting reactant. So the limiting reactant in mixture I must be calcium nitrate. If we do the same for each mixture, we get the limiting reactants as follows. Calcium nitrate is also the limiting reagent in mixture II, III, and IV, while sodium phosphate is the limiting reagent in mixture V.
We are nearly at the end of this complicated calculation. The very last step would be to take the number of moles of product calculated from the limiting reagent in each mixture and use the formula mass equals the number of moles times molar mass to determine the mass of product in each mixture. This calculated mass is the maximum theoretical mass of product, which could form under perfect lab conditions.
Because it is the same product we will be comparing for each mixture, we need not consider the actual molar mass value now. We can just compare the number of moles of product for each mixture. This will indicate to us and give us a sense of the maximum mass of calcium phosphate product in each mixture. The highest number of moles and thus the highest mass of precipitated product is for mixture D. And so the answer is D, mixture number IV.