Answer : The mass of [tex]Ca(OH)_2[/tex] required is 9.3092 grams.
Explanation :
First we have to calculate the mass of solution.
[tex]\text{Mass of solution}=\text{Density of solution}\times \text{Volume of solution}[/tex]
[tex]\text{Mass of solution}=1.065g/mL\times 25.0mL=26.625g[/tex]
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 [tex]\frac{26.625}{100}\times 58.0=15.44g[/tex] of acetic acid
Now we have to calculate the moles of acetic acid.
Molar mass of acetic acid = 60 g/mole
[tex]\text{Moles of }CH_3COOH=\frac{\text{Mass of }CH_3COOH}{\text{Molar mass of }CH_3COOH}=\frac{15.44g}{60g/mole}=0.257moles[/tex]
Now we have to calculate the moles of [tex]Ca(OH)_2[/tex].
The balanced chemical reaction is,
[tex]2CH3CO_2H+Ca(OH)_2\rightarrow Ca(CH_3CO_2)_2+2H_2O[/tex]
From the balanced reaction we conclude that,
As, 2 moles of [tex]CH_3COOH[/tex] react with 1 moles of [tex]Ca(OH)_2[/tex]
So, 0.257 moles of [tex]CH_3COOH[/tex] react with [tex]\frac{0.257}{2}=0.1258[/tex] moles of [tex]Ca(OH)_2[/tex]
Now we have to calculate the mass of [tex]Ca(OH)_2[/tex].
[tex]\text{Mass of }Ca(OH)_2=\text{Moles of }Ca(OH)_2\times \text{Molar mass of }Ca(OH)_2[/tex]
Molar mass of [tex]Ca(OH)_2[/tex] = 74 g/mole
[tex]\text{Mass of }Ca(OH)_2=(0.1258mole)\times (74g/mole)=9.3092g[/tex]
Therefore, the mass of [tex]Ca(OH)_2[/tex] required is 9.3092 grams.