Question

A mordant is a substance that combines with a dye to produced a stable fixed colour in a dyed fabric. calcium Acetate is used as a mordant. It is prepared by the reaction of Acetic acid with calcium hydroxide.
2CH3CO2H + Ca(OH)2 ? Ca(CH3CO2)2 + 2H2O
What mass of Ca(OH)2 is required to react with Acetic Acid in 25.0 ml of a solution having a density of 1.065 g/ mL and containing 58.0% Acetic Acid by mass?

Answer 1

Answer : The mass of $Ca(OH)_2$ required is 9.3092 grams.

Explanation :

First we have to calculate the mass of solution.

$\text{Mass of solution}=\text{Density of solution}\times \text{Volume of solution}$

$\text{Mass of solution}=1.065g/mL\times 25.0mL=26.625g$

Now we have to calculate the mass of acetic acid.

As we are given that the 58.0 % acetic acid by mass. That means, 58.0 grams of acetic acid present in 100 grams of solution.

As, 100 g of solution contains 58.0 g of acetic acid

So, 26.625 g of solution contains $\frac{26.625}{100}\times 58.0=15.44g$ of acetic acid

Now we have to calculate the moles of acetic acid.

Molar mass of acetic acid = 60 g/mole

$\text{Moles of }CH_3COOH=\frac{\text{Mass of }CH_3COOH}{\text{Molar mass of }CH_3COOH}=\frac{15.44g}{60g/mole}=0.257moles$

Now we have to calculate the moles of $Ca(OH)_2$.

The balanced chemical reaction is,

$2CH3CO_2H+Ca(OH)_2\rightarrow Ca(CH_3CO_2)_2+2H_2O$

From the balanced reaction we conclude that,

As, 2 moles of $CH_3COOH$ react with 1 moles of $Ca(OH)_2$

So, 0.257 moles of $CH_3COOH$ react with $\frac{0.257}{2}=0.1258$ moles of $Ca(OH)_2$

Now we have to calculate the mass of $Ca(OH)_2$.

$\text{Mass of }Ca(OH)_2=\text{Moles of }Ca(OH)_2\times \text{Molar mass of }Ca(OH)_2$

Molar mass of $Ca(OH)_2$ = 74 g/mole

$\text{Mass of }Ca(OH)_2=(0.1258mole)\times (74g/mole)=9.3092g$

Therefore, the mass of $Ca(OH)_2$ required is 9.3092 grams.