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What volume of 0.08892$M \mathrm{HNO}_{3}$ is required to react completely with 0.2352 g of potassium hydrogenphosphate?$2 \mathrm{HNO}_{3}(a q)+\mathrm{K}_{2} \mathrm{HPO}_{4}(a q) \longrightarrow \mathrm{H}_{2} \mathrm{PO}_{4}(a q)+2 \mathrm{KNO}_{3}(a q)$

0.03037 $\mathrm{L} \mathrm{HNO}_{3}$

00:47

Sisi G.

Chemistry 101

Chapter 4

Stoichiometry of Chemical Reactions

Chemical reactions and Stoichiometry

Drexel University

Brown University

University of Toronto

Lectures

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to answer this question, we need a balanced chemical reaction. And although a chemical reaction is provided, you may be confused because they have the incorrect chemical formula for phosphoric acid so it does not appear to be balanced. But if we put in the correct chemical formula for phosphoric acid, H three peel four instead of just H two p 04 Then it is balanced with the two in front of the nitric acid and a two in front of the potassium nitrate. So we'll start with the grams of potassium hydrogen phosphate that are reacting and convert them into moles by divided by the molar mass of potassium hydrogen phosphate. Then when we have moles potassium hydrogen phosphate, we can use the two to one relationship to get moles of nitric acid. Knowing the polarity of the nitric acid solution, we can convert the moles nitric acid required into the leaders of the nitric acid solution required by dividing by the polarity of the nitric acid solution. And then if we choose, we can convert the leaders into mill leaders by multiplying by 1000 and we get 30.37 mil. Leaders of nitric acid at a 0.8892 molar concentration are needed to react with all 0.2352 g of potassium hydrogen phosphate.

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