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bixtya [17]
2 years ago
5

An antacid tablet weighing 1.30g was fully neutralized at 42.00 mL(an excess amount) of 0.250MHCl. 10.00 mL of 0.100 M NaOH was

then used to back titrate the excess HCl. How many moles of acid did the antacid neutralize
Chemistry
1 answer:
Yuri [45]2 years ago
7 0

Answer:

0.0095 moles of acid were neutralized by the antiacid

Explanation:

The antiacid is a base that neutralize the acid in stomach. To find the moles of acid neutralized we need to find the moles of acid added initially. This acid is added in excess, then, the moles of NaOH added reacts to neutralize the moles of acid in excess. The difference between initial moles of HCl and moles of NaOH needed to titrate the excess = Moles of HCl that were neturalized by the antiacid as follows:

<em>Moles HCl added:</em>

42.00mL = 0.04200L * (0.250mol/L) = 0.0105 moles HCl

<em>Moles NaOH to titrate the excess:</em>

10.00mL = 0.01000L * (0.10mol/L) = 0.0010 moles NaOH = Moles HCl in excess.

<em>Moles of acid that were neutralized:</em>

0.0105 moles - 0.0010 moles =

<h3>0.0095 moles of acid were neutralized by the antiacid</h3>
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Answer:

The partial pressure of acetic acid is 73.5 torr

Explanation:

Step 1: Data given

Total pressure is 226 torr

mass of acetic acid = 126 grams

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Step 2: Calculate moles of acetic acid

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We can control this by calculating the partial pressure of methanol

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P(methanol) = 0.675 * 226 = 152.55

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