Question

For the following reaction, 30.9 grams of calcium hydroxide are allowed to react with 26.0 grams of hydrochloric acid .

calcium hydroxide(aq) + hydrochloric acid(aq) calcium chloride(aq) + water(l)

What is the maximum amount of calcium 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

Ca(OH)2 + 2HCl ----> CaCl2 + 2H2O

moles of Ca(OH)2 = 30.9gm/74g/mol = 0.42 mol

moles of HCl = 26gm/36.5g/mol = 0.712 mol

According to the balanced reaction 1 mol Ca(OH)2 reacts with 2mol HCl. So, 0.42 mol Ca(OH)2 will react with 2 * 0.42 = 0.84 mol HCl. But the amount of HCl present is 0.712 mol. So, HCl is the limiting reagent and it will be comletely consumed.

Formula of the limiting reagent = HCl

amount of CaCl2 formed = 0.712 mol = 0.712 mol *molar mass of CaCl2 = 79.07 g

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moles of Ca(OH)2 reacted = 0.712/2 = 0.356 mol

Amount of Ca(OH)2 remained = 0.42-0.356 = 0.064 mol

mass of Ca(OH)2 formed = 0.064 mol * 74g/mol = 4.74 gm