Determine the grams of water produced when 250 grams of barium hydroxide react with 125 grams of HCI. Refer to the balanced equation below.

Ba(OH)2 (aq) + 2HCl (aq) --> 2H2O (l) + BaCl2 (aq) (balanced)

Explanation:

Ba(OH)₂ (aq) + 2HCl (aq) --> 2H₂O (l) + BaCl₂ (aq)

171 g                    2 x 36.5 g     2 x 18 g

73 g of HCl reacts with 171 g of Ba(OH)₂

125 g HCl reacts with 171 x 125 / 73 g of Ba(OH)₂

= 292.8 g of  of Ba(OH)₂

Ba(OH)₂  available is 250 g , so it is the limiting reagent .

171 g of Ba(OH)₂ forms 36 g of water

250 g of Ba(OH)₂ forms 36 x 250 / 171 g of water

= 52.63 g of water.