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Answer :
Final answer:
The maximum amount of potassium phosphate that can be formed is 67.03 grams. The formula for the limiting reagent is H3PO4. The amount of excess reagent remaining after the reaction is complete is 48.25 grams of KOH.
Explanation:
To calculate the maximum amount of potassium phosphate that can be formed, we need to determine the limiting reagent. Let's start by calculating the moles of each reactant:
- Potassium hydroxide (KOH):
- Molar mass of KOH = 39.10 g/mol (potassium) + 16.00 g/mol (oxygen) + 1.01 g/mol (hydrogen) = 56.11 g/mol
- Moles of KOH = mass of KOH / molar mass of KOH = 69.9 g / 56.11 g/mol = 1.245 mol
- Phosphoric acid (H3PO4):
- Molar mass of H3PO4 = 1.01 g/mol (hydrogen) + 3(16.00 g/mol) (oxygen) + 31.00 g/mol (phosphorus) = 98.00 g/mol
- Moles of H3PO4 = mass of H3PO4 / molar mass of H3PO4 = 37.9 g / 98.00 g/mol = 0.386 mol
Next, we need to compare the mole ratios of the reactants to the coefficients in the balanced chemical equation to determine the limiting reagent. The balanced equation for the reaction is:
2 KOH(aq) + H3PO4(aq) ⟶ K2HPO4(aq) + 2 H2O(l)
The mole ratio between KOH and H3PO4 is 2:1. Since we have 1.245 moles of KOH and 0.386 moles of H3PO4, we can see that H3PO4 is the limiting reagent because it is present in a smaller amount.
Now, let's calculate the maximum amount of potassium phosphate that can be formed using the stoichiometry of the balanced equation:
- Mole ratio between H3PO4 and K2HPO4 is 1:1
- Maximum moles of K2HPO4 = moles of H3PO4 = 0.386 mol
- Maximum mass of K2HPO4 = moles of K2HPO4 × molar mass of K2HPO4 = 0.386 mol × (39.10 g/mol + 31.00 g/mol + 16.00 g/mol × 4) = 0.386 mol × 174.10 g/mol = 67.03 g
Therefore, the maximum amount of potassium phosphate that can be formed is 67.03 grams.
To determine the formula for the limiting reagent, we can use the mole ratio between KOH and H3PO4. Since the mole ratio is 2:1, the formula for the limiting reagent (H3PO4) is H3PO4.
To calculate the amount of excess reagent remaining, we need to subtract the amount of limiting reagent consumed from the initial amount of excess reagent. Since H3PO4 is the limiting reagent, all of the 37.9 grams of phosphoric acid will be consumed. Therefore, the amount of excess KOH remaining is:
- Initial moles of KOH = 1.245 mol
- Moles of KOH consumed = moles of H3PO4 consumed (based on the stoichiometry of the balanced equation) = 0.386 mol
- Moles of KOH remaining = initial moles of KOH - moles of KOH consumed = 1.245 mol - 0.386 mol = 0.859 mol
- Mass of KOH remaining = moles of KOH remaining × molar mass of KOH = 0.859 mol × 56.11 g/mol = 48.25 g
Therefore, the amount of excess KOH remaining after the reaction is complete is 48.25 grams.
Learn more about calculating the maximum amount of product formed and identifying the limiting and excess reagents here:
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