Thank you for visiting For the following reaction 56 1 grams of potassium hydroxide are allowed to react with 35 9 grams of phosphoric acid text potassium hydroxide aq. This page is designed to guide you through key points and clear explanations related to the topic at hand. We aim to make your learning experience smooth, insightful, and informative. Dive in and discover the answers you're looking for!
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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