8. Carbon dioxide gas is one of the products when Tums (active ingredient: calci

Question

8. Carbon dioxide gas is one of the products when Tums (active ingredient: calcium carbonate) reacts with vinegar (acetic acid). The balanced equation is shown below: Caco, (s) + 2 HC,H,02(aa) CO2 (g) + H2O(/) + Ca(C2H,02)2(aq) a) If 25.0 mL of 0.68 M acetic acid is poured onto a regular strength Tums, containing 500.0 mg of calcium carbonate, what is the limiting reactant? What is the theoretical yield of carbon dioxide, in moles? (4 pts) L.R. = b) What is the theoretical yield of carbon dioxide gas, in liters, at STP? (2 pts)

Explanation / Answer

CaCO3(s) + 2HC2H3O2(aq) --------------> CO2(g) + H2O(l) + Ca(C2H3O2)2(aq)
no of moles of CH3COOH = molarity*volume in L
                        = 0.68*0.025 = 0.017 moles
mass of CaCo3           = 500mg = 0.5g
no of moles of CaCO3    = W/G.M.Wt
                        = 0.5/100 = 0.005 moles
2 moles of Ch3COOH react with 1 moles of CaCo3
0.017 moles of CH3COOH react with = 1*0.017/2 = 0.0085 moles of CaCo3 is required
CaCo3 is limiting reactant
1 moles of CaCO3 react with CH3COOH to gives 1 mole of CO2
0.005 moles of CaCO3 react with Ch3COOH to gives = 1*0.005/2 = 0.0025 moles of CO2
mass of CO2 = no of moles * gram molar mass
             = 0.0025*44 =0.11g of Co2

Theoritical mass of CO2 = 0.11g
b.
1 moles of any gas at STP conditions occupied volume = 22.4L
0.0025 moles of CO2at STP conditions occupied volume = 22.4*0.0025/1 = 0.056L of CO2

theoritical colume of CO2 = 0.056LCO2

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