Chemisty hw question???? Finding mass of Alumium hydroxide?

Question

Q: What mass of solid aluminum hydroxide can be produced when 50.0mL of 0.200 Molarity Al(NO3)3 is added to 200.0 mL of 0.100 Molarity KOH?

A: Please tell me step by step. VERY specific

Answer 1

First write a balanced reaction:

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

From this you can see that 1 mole of Al(NO3)3 will react with 3 moles of KOH to produce 1 mole of Al(OH)3.

You are given 50.0 mL (0.050 L) of 0.2M Al(NO3)3: this containes 0.050*0.2 = 0.010 moles of Al(NO3)3 substance.

You are given 200 ml (0.2 L) of 0.1 M KOH: this contains 0.2*0.1 = 0.02 moles of KOH

A complete reaction needs 3x as much moles of KOH as Al(NO3)3, but 0.010 x 3 > 0.02, so the KOH is the limiting reagent (there is too much Al(NO3)3 so there will be some left over after the reaction). Thus we only need to consider the amount of KOH used (0.02 moles).

For every 3 moles of KOH, 1 moles of Al(OH)3 is produced. This reaction will therefore produce (1/3) * 0.02 moles of Al(OH)3, or 0.00667 moles of AL(OH)3. The molecular mass of Al(OH)3 is 78 grams/mole, so the mass is 0.00667 * 78 = 0.52 grams