25.00mL of 0.125M aluminum chloride solution is mixed with 25.00mL of 0.100M sodium hydroxide solution. What mass of aluminum hydroxide will be formed?

Question

Help!! Chemistry word problems. Solution Stoichiometry.

Answer 1

Hi Bee!

 

1) Find limiting reagent (aka limiting reactant):

 

0.025 L AlCl3 x 0.125 mol/L AlCl3 x 1 mol Al(OH)3/1 mol AlCl3 = 0.003125 mol Al(OH)3

 

vs.

 

0.025 L NaOH x 0.1 mol/L NaOH x 1 mol Al(OH)3/3 mol NaOH = 0.00083 mol Al(OH)3

 

So NaOH is the limiting reagent

 

2) Since NaOH runs out first, 0.00083 mol of Al(OH)3 is formed.

 

3) Divide moles by the total volume (50 mL):

 

0.00083 mol/0.05 L = 0.0167 M Al(OH)3  

 

3 sig figs!

 

Hope this helps!

 

-- Sarah