Question

For the following reaction, 22.9 grams of hydrochloric acid are allowed to react with 57.2 grams of barium hydroxide. hydrochloric acid (aq) + barium hydroxide (aq) —barium chloride (aq) + water (1) What is the maximum amount of barium chloride that can be formed? grams What is the FORMULA for the limiting reagent? What amount of the excess reagent remains after the reaction is complete? grams

Answer 1

the balanced equation is as follows

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

no of moles = mass / molar mass

for HCl = 22.9 g / 36.5 g/mol => 0.6274 mol

for Ba(OH)2 = 57.2g / 171.34 g/mol => 0.3338 mol

2 mol HCl ------------> 1 mol BaCl2

0.6274 mol ----------------->?

=> 0.6274 * 1 / 2

=> 0.3137 mol

1 mol Ba(OH)2 ----------------> 1 mol BaCl2

0.3338 mol -----------------------> 0.3338 mol

limiting reactant is HCl

theoretical yield => 0.3137 mol * 208.23 g/mol => 65.3 grams

excess reactant moles => 0.02014 moles

excess reactant mass after reaction => 0.02014 mol * 171.34 g/mol => 3.45 g

answers =>

BaCl2 mass = 65.3 grams

limiting reactant = HCl

excess reactant mass after reaction 3.45 g