2. In the following reaction 30 grams of sodium carbonate Na CO, and 38 grams of

Question

2. In the following reaction 30 grams of sodium carbonate Na CO, and 38 grams of HCl were used. Na CO,+ 2 HC2 NaCI+ CO, + HO a) Calculate how many grams of sodium chloride will be formed. b) Calculate how many grams of water will be formed. c) Calculate how many grams of carbon dioxide will be formed. 3. In the following reaction 702 grams of aluminum carbonate was used to react with 196 grams phosphoric acid. Al(CO,),2H,PO, B 2AIPo,+ 3CO, +3HO d) Calculate how many grams of aluminum phosphate will be formed e) Calculate how many grams of water will be formed. f) Calculate how many grams of carbon dioxide will be formed.

Explanation / Answer

2.   Na2Co3 + 2HCl -------------> 2NaCl + Co2 + H2O
     no of moles of Na2Co3 = W/G.M.Wt
                             = 30/106 = 0.283moles
     no of moles of HCl    = W/G.M.Wt
                           = 38/36.5 = 1.04 moles
2 moles of HCl react with 1 mole of Na2Co3
1.04 moles of HCl react with = 1*1.04/2   = 0.52 moles of Na2Co3
   Na2Co3 is limiting reactant
1 mole of Na2Co3 react with HCl to gives 2 moles of NaCl
0.283 moles of Na2Co3 react with HCl to gives = 2*0.283/1   = 0.566 moles of NaCl
mass of NaCl = no of moles * gram molar mass
               = 0.566*58.5 = 33.111g
c.1 moles of Na2Co3 react with HCl to gives 1 mole of CO2
0.283 moles of Na2Co3 react with HCl to gives = 1*0.283/1 = 0.283 moles of Co2
mass of CO2 = no of moles * gram molar mass
              = 0.283*44 = 12.452g

b.1 moles of Na2Co3 react with HCl to gives 1 mole of H2O
0.283 moles of Na2Co3 react with HCl to gives = 1*0.283/1 = 0.283 moles of H2O
mass of CO2 = no of moles * gram molar mass
              = 0.283*18 = 5.094g

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