H3PO4 + 3KOH = K3PO4 + 3 H2O What would be the percent yield if you reacted 49.0g of H3PO4 with 112g of potassium hydroxide, and collected 25.0g of water?
To calculate the percent yield, you need to compare the actual yield (the amount of water you collected) with the theoretical yield (the amount of water that would be produced based on the balanced equation).
First, let's calculate the molar mass of H3PO4:
H3PO4: 3(1.00784 g/mol) + 1(15.999 g/mol) + 4(31.9988 g/mol) = 97.994 g/mol
Next, calculate the molar mass of KOH:
KOH: 1(39.0983 g/mol) + 1(1.00784 g/mol) + 16.00 g/mol = 56.105 g/mol
Now, let's calculate the number of moles of H3PO4:
moles of H3PO4 = mass of H3PO4 / molar mass of H3PO4
moles of H3PO4 = 49.0 g / 97.994 g/mol = 0.500 mol
Next, calculate the number of moles of KOH:
moles of KOH = mass of KOH / molar mass of KOH
moles of KOH = 112 g / 56.105 g/mol = 1.997 mol
According to the balanced equation, the ratio between H3PO4 and H2O is 1:3. This means that for every mole of H3PO4, 3 moles of H2O should be produced.
The theoretical yield of water can be calculated based on the moles of H3PO4:
moles of H2O = 3 * moles of H3PO4 = 3 * 0.500 mol = 1.500 mol
Now, calculate the mass of the theoretical yield of water:
mass of H2O = moles of H2O * molar mass of H2O
mass of H2O = 1.500 mol * 18.015 g/mol = 27.023 g
Finally, calculate the percent yield:
percent yield = (actual yield / theoretical yield) * 100
percent yield = (25.0 g / 27.023 g) * 100 = 92.5%
Therefore, the percent yield in this reaction is 92.5%.