Intro to chemistry

Question

Consider the reaction:

P₂O₅(s) + 3 H₂O(l) → 2 H₃PO₄(aq)

If 12.32 g P₂O₅ is mixed with 16.76 g H₂O, calculate the theoretical yield (g) of H₃PO₄ produced by the reaction. Need this answer and how you got it asap.

Answer 1

Theoretical yield = 17.01 g of H3PO4. P2O5 is the limiting reactant; the water is in large excess. Working and the reasoning below.

Step 1: convert both masses to moles.

M(P2O5) = 2(30.97) + 5(16.00) = 141.94 g/mol
M(H2O)  = 2(1.008) + 16.00  = 18.02 g/mol

n(P2O5) = 12.32 g / 141.94 g/mol = 0.08680 mol
n(H2O)  = 16.76 g / 18.02 g/mol  = 0.9301 mol

Step 2: find the limiting reactant. Do not compare the moles directly, and never compare the masses. Divide each amount by its coefficient in the balanced equation, then the smaller number is the limiting one:

P2O5:  0.08680 / 1 = 0.08680   (smallest -> LIMITING)
H2O :  0.9301  / 3 = 0.3100

Another way to see the same thing: 0.08680 mol of P2O5 needs 3 x 0.08680 = 0.2604 mol of water, and you have 0.9301 mol. There is over three and a half times more water than the reaction can use, so the P2O5 runs out first and it alone sets the yield.

Step 3: use the mole ratio from the balanced equation. One P2O5 gives two H3PO4, so multiply by 2, not by 1:

n(H3PO4) = 2 x 0.08680 = 0.1736 mol
M(H3PO4) = 3(1.008) + 30.97 + 4(16.00) = 97.99 g/mol
mass     = 0.1736 mol x 97.99 g/mol = 17.01 g

Check it with conservation of mass, which is free and catches almost every slip here. All of the limiting reactant and only part of the water end up in the single product, so the product mass must equal the mass of P2O5 plus the mass of water actually consumed:

water consumed = 0.2604 mol x 18.02 = 4.69 g
12.32 g + 4.69 g = 17.01 g   (matches)

Left over at the end: 0.9301 - 0.2604 = 0.6697 mol of water, which is 12.07 g of unreacted H2O sitting in the flask with your product.

Two things worth taking away from this problem.

First, "theoretical yield" means the maximum the stoichiometry allows if the reaction goes to completion and you lose nothing in transfer, filtration or drying. Whatever you actually weigh out is the actual yield, and percent yield = (actual / theoretical) x 100. So 17.01 g is the ceiling for this experiment, not a prediction of what will be on the balance.

Second, the reason the water is deliberately in such huge excess is practical rather than arithmetic. P2O5 is one of the most aggressive drying agents in the lab, and it attacks water violently and exothermically. Running it into a large volume of water rather than the other way around keeps the heat under control, and the leftover water is simply the solvent the phosphoric acid ends up dissolved in. Excess reactant is very often the solvent, which is why "how much is left over" is a sensible thing for a question to ask.