What mass of solid aluminum hydroxide can be produced when 50.0 mL of 0.200M Al(No3)3 is added to 200 Ml of a 0.100M KoH
trying to figure this out myself
To calculate the mass of solid aluminum hydroxide produced, you can use the stoichiometry of the balanced chemical equation for the reaction between aluminum nitrate (Al(NO3)3) and potassium hydroxide (KOH).
First, let's write the balanced chemical equation for the reaction:
2 Al(NO3)3 + 6 KOH -> 2 Al(OH)3 + 6 KNO3
From the balanced equation, we can see that 2 moles of aluminum nitrate react with 6 moles of potassium hydroxide to produce 2 moles of solid aluminum hydroxide.
Now, let's find the number of moles of aluminum nitrate and potassium hydroxide present in the solution:
Moles of Al(NO3)3 = volume (in L) x concentration (in mol/L) = 0.050 L x 0.200 mol/L = 0.010 mol
Moles of KOH = volume (in L) x concentration (in mol/L) = 0.200 L x 0.100 mol/L = 0.020 mol
According to the balanced equation, the reaction takes place in a 1:3 molar ratio between Al(NO3)3 and KOH. Therefore, the limiting reactant is Al(NO3)3 since there are fewer moles of it.
Now, let's calculate the moles of solid aluminum hydroxide produced:
Moles of Al(OH)3 = (0.010 mol Al(NO3)3) x (2 mol Al(OH)3 / 2 mol Al(NO3)3) = 0.010 mol Al(OH)3
Finally, to find the mass of aluminum hydroxide produced, we need to multiply the number of moles by its molar mass, which is approximately 78 g/mol:
Mass of Al(OH)3 = (0.010 mol Al(OH)3) x (78 g/mol) = 0.78 g
Therefore, approximately 0.78 grams of solid aluminum hydroxide can be produced when 50.0 mL of 0.200M Al(NO3)3 is added to 200 mL of 0.100M KOH.