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Answer :
The maximum amount of potassium phosphate that can be formed is 77.8 grams.
- The balanced chemical equation for the reaction is:
[tex]\[3\text{KOH} (aq) + \text{H}_3\text{PO}_4 (aq) \rightarrow \text{K}_3\text{PO}_4 (aq) + 3\text{H}_2\text{O} (l)\][/tex]
- To find the maximum amount of potassium phosphate formed, we first need to determine the limiting reactant.
- Calculate the moles of each reactant:
- Moles of [tex]\(\text{KOH}\): \(\frac{56.1 \text{ g}}{56.1 \text{ g/mol}} = 1.0 \text{ mol}\)[/tex]
- Moles of [tex]\(\text{H}_3\text{PO}_4\): \(\frac{35.9 \text{ g}}{98.0 \text{ g/mol}} = 0.366 \text{ mol}\)[/tex]
- Since the stoichiometry of the reaction is 3 moles of [tex]\(\text{KOH}\)[/tex] to 1 mole of [tex]\(\text{H}_3\text{PO}_4\), \(\text{H}_3\text{PO}_4\)[/tex] is the limiting reactant.
- Calculate the theoretical yield of [tex]\(\text{K}_3\text{PO}_4\)[/tex]:
[tex]\[0.366 \text{ mol} \times \frac{1 \text{ mol K}_3\text{PO}_4}{1 \text{ mol H}_3\text{PO}_4} \times 212.27 \text{ g/mol} = 77.8 \text{ g}\][/tex]
- Therefore, the maximum amount of potassium phosphate that can be formed is 77.8 grams.
Complete Question:
For the following reaction, 56.1 grams of potassium hydroxide are allowed to react with 35.9 grams of phosphoric acid.
potassium hydroxide ([tex]\(aq\)[/tex]) + phosphoric acid ([tex]\(aq\)[/tex]) ⟶ potassium phosphate ([tex]\(aq\)[/tex]) + water ([tex]\(l\)[/tex])
A. What is the maximum amount of potassium phosphate that can be formed?
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