Question

A sample of 5.02 g of solid calcium hydroxide is added to 32.5 mL of 0.440 M aqueous hydrochloric acid Write the balanced chemical equation for the reaction. Physical states are optional chemical equation: What is the limiting reactant? hydrochloric acid calcium hydroxide How many grams of salt are formed after the reaction is complete? mass of salt: How many grams of the excess reactant remain after the reaction is complete? excess reactant remaining:

Answer 1

given

5.02 g Ca(OH)2 and 32.5 mL of 0.440 M HCl

So

Reaction is                                                                              

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

Balancing the equation

Consider Ca

It is balanced

Consider Cl

In the product side 2 Cl so multiply HCl by 2 on reactant side

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

Consider H

In the reactant side 4 H so multiply H2O by 2 on product side

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

Consider O

It is balanced

So

balanced equation is

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

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Lets calculate no of moles for each reactant

no of moles of Ca(OH)2 = Mass of Ca(OH)2 / Molar Mass of Ca(OH)2

= 5.02 g/74.093 g/mol

= 0.068 moles

no of moles of HCl = Molarity * Volume in L

= 0.440 moles/L * (32.5/1000) L

= 0.0143 moles

From the reaction we can say that

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

So

0.068 moles of Ca(OH)2 will react with

0.068 moles of Ca(OH)2 * (2 moles of HCl/1 mole of Ca(OH)2)

= 0.136 moles of HCl

But we have 0.0143 moles of HCl which is less

hence

HCl is the limiting reactant

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Now

From the reaction we can say that

2 moles of HCl produces 1 mole of CaCl2

So

0.0143 moles of HCl will produce

0.0143 moles of HCl * (1 mole of CaCl2/2 moles of HCl)

= 0.00715 moles of CaCl2

Mass of CaCl2 formed = moles of CaCl2 * Molar moles of CaCl2

= 0.00715 moles * 110.98 g/mol

= 0.794 g CaCl2 salt is formed

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Now

From the reaction we can say that

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

So

0.0143 moles of HCl will react with

0.0143 moles of HCl*(1 mole of Ca(OH)2/2 moles of HCl)

= 0.00715 moles Ca(OH)2

So

Remaining moles of Ca(OH)2 = 0.068 moles - 0.00715 moles

= 0.061 moles

Mass of remaining Ca(OH)2 = (moles of Ca(OH)2) * Molar Mass of Ca(OH)2

= 4.52 grams of excess reactant (Ca(OH)2) is remaining