10.0 grams of barium chloride (208.3g/mol) is reacted with excess potassium phosphate and 7.22 of grams of barium phosphate (601.8 g/mol) is isolated. calculate the theoretical yield of barium phosphate
I suspect a balanced equation is the place to start...
3BaCl2 + 2K3PO4 >>> Ba3(PO4)2 + 6KCl
moles barium phosphate: 1/3 *moles barium chloride
calcuate the moles barium chloride, divide by three, then that is the moles of barium phosphate, convert that to grams of barium phosphate.
To calculate the theoretical yield of barium phosphate, we need to determine the limiting reactant in the reaction between barium chloride and potassium phosphate. The limiting reactant is the one that is completely consumed in the reaction and determines the maximum amount of product that can be formed.
First, we need to determine the number of moles of barium chloride and potassium phosphate. We can use the formula: moles = mass / molar mass.
Moles of barium chloride = 10.0 g / 208.3 g/mol = 0.048 moles
Moles of barium phosphate = 7.22 g / 601.8 g/mol = 0.012 moles
Next, we compare the moles of barium chloride and potassium phosphate in the reaction equation:
3 BaCl2 + 2 K3PO4 → Ba3(PO4)2 + 6 KCl
From the balanced equation, we see that the ratio of moles between barium chloride and barium phosphate is 3:1. Therefore, for every 3 moles of barium chloride, we would expect 1 mole of barium phosphate.
Since the stoichiometric ratio is 3:1, the calculated moles of barium chloride (0.048 moles) will give us a theoretical yield of:
= 0.048 moles x (1 mole Ba3(PO4)2 / 3 moles BaCl2)
= 0.048 moles x 1/3
= 0.016 moles of barium phosphate
Finally, to find the mass of barium phosphate, we can use the formula: mass = moles x molar mass.
Mass of barium phosphate = 0.016 moles x 601.8 g/mol
= 9.63 grams
Therefore, the theoretical yield of barium phosphate is 9.63 grams.