Answer:
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.
