Chemistry help... Please explain?

Question

plz help with any of these problems

A 2.29 g sample of an unknown acid is dissolved in 1.0 liter of water. A titration required 25.0 ml of 0.500 M NaOH to completely react with all the acid present. What is the molar mass of the acid?

What mass of aluminum hydroxide is produced when 50.0 ml of 0.200 M Al(NO3)3 reacts with 200.0 ml of 0.100 M KOH?

Answer 1

First question)

It makes a difference whether or not the acid is monoprotic or diprotic. Assume it is monoprotic (one replaceable hydrogen). Then, 1 mol of acid reacts with 1 mol of base.

Find the moles of NaOH:

M = mol / L

mol = M x L = 0.500 mol/L x 0.0250 L = 0.0125 mol NaOH

0.0125 mol NaOH x (1 mol acid / 1 mol NaOH) = 0.0125 mol acid

Molar mass = grams/mole = 2.29 g / 0.0125 mol = 183 g/mol

Second question)

The reaction is:

Al(NO3)3 + 3KOH ----->  Al(OH)3 + 3KNO3

Find the moles of each reactant to determine the limiting reagent:

M = mol / L

mol = M X L = 0.200 mol/L x 0.0500 L = 0.0100 mol Al(NO3)3

mol = M X L = 0.100 mol/L x 0.2000 L = 0.0200 mol KOH

From the balanced equation, 1 mol Al(NO3)3 reacts with 3 mol KOH. KOH is the limiting reagent because there is only twice as much KOH, but three times as much is needed. Use KOH to determine the yield of Al(OH)3.

From the balanced equation, 3 mol KOH produces 1 mol Al(OH)3.

0.0200 mol KOH x (1 mol Al(OH)3 / 3 mol KOH) = 0.00667 mol Al(OH)3

The molar mass of Al(OH)3 is 78.0 g/mol

0.00667 mol x (78.0 g / 1 mol) = 0.520 g Al(OH)3