If 42.7 ml of a 0.208 m hcl solution is needed to neutralize a solution of ca(oh)2, how many grams of ca(oh)2 must be in the solution? ca(oh)2(aq) + 2hcl  cacl2(aq) + 2h2o(l) g

The mass of calcium hydroxide that are present in the solution is 0.3256 grams.

Explanation:

To calculate the moles of a solute, we use the equation:

We are given:

Volume of hydrochloric acid = 42.7mL = 0.0427 L   (Conversion factor: 1 L = 1000 mL)

Molarity of the solution = 0.208 moles/ L

Putting values in above equation, we get:

For the given chemical reaction:

By Stoichiometry of the reaction:

2 moles of hydrochloric acid reacts with 1 mole of calcium hydroxide.

So, 0.0088 moles of hydrochloric acid will react with = of calcium hydroxide.

To calculate the mass of calcium hydroxide, we use the equation:

Moles of calcium hydroxide = 0.0044 moles

Molar mass of calcium hydroxide = 74 g/mol

Putting values in above equation, we get:

Hence, the mass of calcium hydroxide that are present in the solution is 0.3256 grams.

it's filtration, but this is an assumption so my apologies if i got this