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For the following reaction, 0.557 moles of potassium hydroxide are mixed with 0.224 moles of phosphoric acid. potassium hydroxide(aq)+phosphoric acid(aq)= potassium phosphate(aq)+water(l)   What is the formula for the limiting reagent? What is the maximum amount of potassium phosphate that can be produced?

For the following reaction, 0.557 moles of potassium hydroxide are mixed with 0.224 moles of phosphoric acid. potassium hydroxide(aq)+phosphoric acid(aq)= potassium phosphate(aq)+water(l)   What is the formula for the limiting reagent? What is the maximum amount of potassium phosphate that can be produced?

Chemistry
Chemistry
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ISBN:9781305957404
Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
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For the following reaction, 0.557 moles of potassium hydroxide are mixed with 0.224 moles of phosphoric acid.

potassium hydroxide(aq)+phosphoric acid(aq)= potassium phosphate(aq)+water(l)
 

What is the formula for the limiting reagent?

What is the maximum amount of potassium phosphate that can be produced?

Expert Solution
Step 1

(a) The reaction can be given as:

3KOH(aq)+H3PO4(aq)→K3PO4(aq)+3H2O(l)

For the reaction, it can be seen that,

3 mol of potassium hydroxide reacts with 1 mol of phosphoric acid.

0.557 mol of potassium hydroxide reacts with x mol of phosphoric acid.

Therefore,

x=1×0.5573x=0.185 mol

But, the number of moles of phosphoric acid given is 0.224 moles out of which 0.185 moles are used 

Therefore, the remaining amount of phosphoric acid is:

0.224-0.185=0.039 moles

Therefore, 0.039 moles of phosphoric acid is in excess.

Therefore, phosphoric acid is the excess reagent and potassium hydroxide is a limiting reagent. 

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