Solid calcium bicarbonate (Ca(HCO3)2) reacts with aqueous hydrochloric acid (HCl

Question

Solid calcium bicarbonate (Ca(HCO3)2) reacts with aqueous hydrochloric acid (HCl) to form aqueous calcium chloride, liquid water, and carbon dioxide gas.

(a) Write the balanced molecular equation, complete ionic equation, and net ionic equation for the process. [Hint: Don’t forget the physical states of the reactants and products.]

Balanced Molecular Equation:

Complete Ionic Equation:

Net Ionic Equation:

(b) A CHEM 104 student is trying to carry out this reaction in the laboratory. What is the minimum amount of 0.2102 M hydrochloric acid (in milliliters) required to completely react with 8.362 g calcium bicarbonate?

Explanation / Answer

Answer: According to question : The balanced equation is ,

Ca(HCO3)2(s) + 2HCl(aq) ---> 2H2O(l) + 2CO2(g) + CaCl2(aq)

It is the balanced equation.

Now the ioniv equation is : Ca2+ [S] + 2Cl- (aq) -------> Ca2+(aq) + 2Cl-(aq) + 2CO2(g) + 2H2O(l)

2] Now , According to the given informations , Here we know that 1 mol of calcium bicarbonate is required 2 mol of HCl

Hence , 162.1146 g of calcium bicarbonate required = 72.92 g of HCl

hence 8.362 g required = 72.92 * 8.362 / 162.1146 g of HCl = 3.76 g of HCl

Now the number of moles of Hcl = mass / molar mass = 3.76 /36.46 = 0.103 mol

Now , according to question here we have given the molarity of Hcl = 0.2102 M = numer of moles / volume [in l]

hence volume = 0.103 /0.2102 = 0.490 L = 490 ml

Hence the required volume is 490ml .

Thank you :)

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