Moles of HCl in this reaction
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# Moles of HCl in this reaction

[From: ] [author: ] [Date: 11-07-07] [Hit: ]
The student boils the mixture and then allows it to cool. Finally, the student adds bromophenol blue indicator to the mixture, which turns yellow, and then adds 11.72 mL of 0.......
A student dissolves 0.326 g. of a powdered antacid in 32.36 mL of 0.1034 M HCl. The student boils the mixture and then allows it to cool. Finally, the student adds bromophenol blue indicator to the mixture, which turns yellow, and then adds 11.72 mL of 0.1506 M NaOH to turn the solution from yellow to blue.

a.) How many moles of HCl reacted with the antacid? How many equivalents of antacid are present in the sample?

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First, find out how many moles of HCl you started with. 0.1034 M is 0.1034 moles per liter. Multiply that my the number of liters you have (32.36/1000) so 0.1034 x (32.26/1000) = 0.003346 moles HCl you started with.

Next how many moles of HCl were neutralized by the NaOH? Same calculation different values:
0.1506 x (11.72/1000) = 0.001765 moles of NaOH. Since NaOH and HCl each have one equivalent, the moles are equal (that is, 0.001765 miles of NaOH will neutralize 0.001765 moles of HCl).

So the moles neutralized by the antacid is 0.003346 - 0.001765 = 0.001581 moles HCl reacted with the antacid. And, since HCl has one equivalent per mole, the antacid had 0.001581 equivalents.
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