A sample of 8.51 g of solid calcium hydroxide is added to 26.5 mL of 0.340 M aqu

Question

A sample of 8.51 g of solid calcium hydroxide is added to 26.5 mL of 0.340 M aqueous hydrochloric acid. Enter the balanced chemical 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 O hydrochloric acid How many grams of salt are formed after the reaction is complete? Number How many grams of the excess reactant remain after the reaction is complete? Number

Explanation / Answer

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

balance

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

b)

mmol of Ca(OH)2 = mass/MW = 8.51/74.0927 = 0.1148 mol of Ca(OH)2

mol of HCl = MV = (0.34)(26.5*10^-3) = 0.00901 mol of HCl

clearly, there is excess base, so

HCl is limiting

c)

mass of salt formed --> 0.00901 mol of HCl --> 0.00901*1/2

0.004505 mol of CaCl2 forms

mass = mol*MW = 0.004505*110.98 = 0.4999 = 0.50 g of CaCl2

d9)

excess left -->

mol of Ca(OH)2 =0.1148 -  0.00901 = 0.10579 mol of base left

mass = mol*MW = 0.10579*74.1 = 7.839 g of Ca(OH)2 left

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