Answer:
Limiting reactant: P₂O₅
Grams of H₃PO₄ : 2.94 g
Excess reagent: 13.58 g of H₂O
Explanation:
So first thing you need to do is look at your balanced equation:
2P₂O₅ + 6H₂O → 4H₃PO₄
Given this equation, you will need:
2 moles of P₂O₅ to produce 4 moles of H₃PO₄
So the ratio would be:
[tex]\dfrac{2\ moles\ of\ P_2O_5}{4\ moles\ of\ H_3PO_4}[/tex]
6 moles of H₂O to produce 4 moles of H₃PO₄
So the ration would be:
[tex]\dfrac{6\ moles\ of\ H_2O}{4\ moles\ of\ H_3PO_4}[/tex]
These are what we call theoretical yields.
Now since the reactants given are in grams, we need to find out how many moles are there, given the mass. We just add up the molecular mass of each element in the reactants to determine that:
4.8 g of P₂O₅:
P = 2 x 30.97 g/mole = 61.94 g/mole
O = 5 x 16.00 g/mole = 80.00 g/mole
141.94 g/mole
We then determine how many moles given the mass:
[tex]4.8\ g\ of\ P_2O_5 \ \times \ \dfrac{1\ mole\ of\ P_2O_5}{141.94g} = 0.03\ moles\ of P_2O_5[/tex]
15.2 g of H₂O
H = 2 x 1.00 g/mole = 2.00 g/mole
O = 1 x 16.00 g/mole = 16.00 g/mole
18.00 g/mole
[tex]15.2\ g\ of\ H_2O \ \times \ \dfrac{1\ mole\ of\ H_2O}{18.00g} = 0.84\ moles\ of H_2O[/tex]
Next we determine how many grams of the product are there in one mole:
H = 3 x 1.00 = 3.00 g/mole
P = 1 x 30.97 = 30.97 g/mole
O = 4 x 16.00 = 64.00 g/mole
97.97 g/mole
So now we compare that to the theoretical yield to determine how many moles of the product we can produce and convert that into grams given the following grams of reactant.
4.8 g of P₂O₅:
[tex]0.03\ moles\ of\ P_2O_5\times\ \dfrac{4\ moles\ of\ H_3PO_4}{4\ moles\ of\ P_2O_5}\times \dfrac{97.97g\ of\ H_3PO_4}{1\ mole \ of\ H_3PO_4} = 2.94g\ of\ H_3PO_4[/tex]
15.2 g of H₂O
[tex]0.84\ moles\ of\ H_2O\times\ \dfrac{4\ moles\ of\ H_3PO_4}{6\ moles\ of\ H_2O}\times \dfrac{97.97g\ of\ H_3PO_4}{1\ mole \ of\ H_3PO_4} = 54.86g\ of\ H_3PO_4[/tex]
Which ever produces less is the limiting reactant, and what produces the more is the excess reagent. And because the reaction will stop when one reactant is used up, the limiting reactant is what will determine how much of the product is produced.
Now to determine how much of the excess reactant will be left, we just need to compare the limiting reactant to the ratio between the two reactants used and convert it into grams. So here we go:
[tex]0.03\ moles\ of\ P_2O_5 \times \dfrac{6\ moles\ of H_2O}{2\ moles\ of\ P_2O_5}\times\dfrac{18.00g}{1\ mole\ of\ H_2O}= 1.62g\ of\ H_2O[/tex]
This means that when you use up the limiting reactant, you will use 1.62 g of H₂O.
So we subtract that from what we have available:
15.2 grams of H₂O -- 1.62 g of H₂O = 13.58 g of H₂O
