Question

A sample of 8.90 g of solid calcium hydroxide is added to 26.0 mL of 0.190 M aqueous hydrochloric acid. Enter the balanced chemical equation for the reaction. Physical states are optional and not graded.

A) What is the limiting reactant?

B) How many grams of salt is formed after the reaction is complete?

C) How many grams of the excess reaction remain after the reaction is complete?

Answer 1

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

Molar mass of Ca(OH)2 = 40 + 2*(1+16) = 74 gm/mol

Number of moles of Ca(OH)2 = 8.90/74 = 0.12027 moles

Number of moles of HCl = 26/1000 * 0.190 = 0.004992 moles

A) HCl is the limiting reagent in the reaction

B) 2 moles of HCl will give 1 mole of Ca(Cl)2 salt

Number of moles of CaCl2 formed = 0.004992/2 = 0.002496 moles

Molar mass of CaCl2 = 40 + 2 * 35.5 = 111 gm/mol

Mass of CaCl2 formed = 0.002496 moles * 111 gm/mol = 0.277056 grams

c) Excess Ca(OH)2 left = 0.12027 - 0.002496 = 0.117774 moles

Mass of Ca(OH)2 left = 0.11774 moles * 74 gm/mol = 8.715276 grams