A sample of solid calcium hydroxide is added to 21.0 mL of 0.300 M aqueous hydro

Question

A sample of solid calcium hydroxide is added to 21.0 mL of 0.300 M aqueous hydrochloric acid. Enter the balanced equation for the reaction. Physical states are optional.
What is the limiting reactant?
How many grams of salt are formed after the reaction is complete?
How many grams of the excess reactant remain after the reaction is complete? Question 40 of 55 Map Sapling Learning A sample of 7.44 g of solid calcium hydroxide is added to 21.0 mL of 0.300 M aqueous hydrochloric acid. Enter the balanced chenical equation for the reaction. Physical states are optional and not graded. Tip: If you need to clear your work and reset the equation, click the button that looks like two arrows. What is the limiting reactant? O calcium hydroxide hydrochloric acid How many grams of salt are formed after the reaction is compiete? Number How many grams of the excess reactant remain after the reaction is complete? Number O Previous Next Save And Exit

Explanation / Answer

The balance chemical reaction is

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

Thus 1 mole Ca(OH)2 reacts with 2 moles of HCl to form 1 mole CaCl2.

Now molar mass of Ca(OH)2 is 74.093 g/mol while molar mass of HCl is 36.46 g/mol and molar mass of CaCl2 is 110.98 g/mol. Hence, 74.093 g of Ca(OH)2 reacts with 72.92 g of HCl to form 110.98 g/mol of CaCl2

Now we know molarity = moles of solute/volume of solution in litre.

Hence, amount of HCl in 21.0 mL of 0.30 M of HCl = 6.3 X 10-3 mol = 0.2297 g of HCl.

Since 72.92 g of HCl reacts with 74.093 g of Ca(OH)2, 0.2297 g HCl will react with 0.2334 g Ca(OH)2.

Since all the HCl is consumed and Ca(OH)2 remain in excess, the limiting reactant is hydrochloric acid.

As 72.92 g of HCl reacts with 74.093 g of Ca(OH)2 to form 110.98 g Salt (CaCl2), 0.2297 g HCl will react with 0.2334 g Ca(OH)2 to form 0.3496 g CaCl2.

The excess reactant is = 7.44- 0.2334 g = 7.2066 g Ca(OH)2.

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