What volume, in mL, of 0.1234 M HCl is needed to neutralize 2.00 g Ca(OH)2?



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Posted on (Answer #1)

First let us make a balanced chemical equation for this reaction.

`2 HCl + Ca(OH)_2 -> CaCl_2 + 2 H_2O`

To solve this problem, we should first compute how many moles of Ca(OH)2 are needed to be reacted. Using the above equation, we can determine the amount of moles of HCl and eventually the volume of the HCl solution.

`2.00 grams Ca(OH)_2 * (1 mol e Ca(OH)_2)/(74.093 grams Ca(OH)_2) * (2 mol es HCl)/(1 mol e Ca(OH)_2)`

= 0.05399 moles HCl

From molarity:

`Molarity = (mol es solute)/(volume of solution (L))`

`Volume of solution = (0.05399 mol es HCl)/(0.1234 (mol es)/(L))`

Volume of solution = 0.4375 L = 438 mL



Note: 1000mL = 1L


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