A sample of 957 g of solid calcium hydroxide is added to 295

A sample of 9.57 g of solid calcium hydroxide is added to 29.5 mL of 0.280 M aqueous hydrochloric acid. Enter the balanced chemical equation for the reaction. Physical states are optional and not graded. Ca(OH)2+2HCl CaCl2 +2 H,O 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 complete? Number 45843 ow many grams of the excess reactant remain after the reaction is complete? Number 20.99

Solution

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

no of moles of Ca(OH)2    = W/G.M.Wt

                                         = 9.57/74    = 0.13 moles of Ca(OH)2

no of moles of HCl   = molarity* volume in L

                                 = 0.28*0.0295   = 0.00826 moles of HCl

1 moles of Ca(OH)2 react with 2 moles of HCl

0.13 moles of Ca(OH)2 react with = 2*0.13/1   = 0.26 moles of HCl is required

HCl is limiting reactant

2 moles of HCl react with Ca(OH)2 to gives 1 moles of CaCl2

0.00826 moles of HCl react with Ca(OH)2 to gives = 1*0.00826/2   = 0.00413 moles of CaCl2

mass of CaCl2 = no of moles * gram molar mass

                         = 0.00413*111   = 0.4584g of CaCl2

2 moles of HCl react with 1 mole of Ca(OH)2

0.00826 moles of HCl react with = 1*0.00826/2   = 0.00413 moles of Ca(OH)2

The no of moles of excess remains after complete the reaction = 0.13-0.00413   = 0.12587moles of Ca(OH)2

The amount of excess remains after complete the reaction = no of moles * gram molar mass

                                                                                            = 0.12587*74   = 9.3g