How many moles and liters of gas would be produced by reacting77.5 g of NaHCO 3

Question

How many moles and liters of gas would be produced by reacting77.5 g of NaHCO3 with excess vinegar? Assume thereaction releases all the gas and that the reaction takes place at22 degrees C and 1 atm. NaHCO3 (s) + CH3COOH (aq) =>CO2 (g) + H2O (l) +NaC2H3O2 (aq) How many moles and liters of gas would be produced by reacting77.5 g of NaHCO3 with excess vinegar? Assume thereaction releases all the gas and that the reaction takes place at22 degrees C and 1 atm. NaHCO3 (s) + CH3COOH (aq) =>CO2 (g) + H2O (l) +NaC2H3O2 (aq)

Explanation / Answer

We Know that :       The given reaction is :         NaHCO3 (s) + CH3COOH (aq) =>CO2 (g) + H2O (l) +NaC2H3O2 (aq)         number of molesof NaHCO3 = 77.5 g / 84 g /mol                                                          = 0.9266 mol       From the balanced equatin it isclear that 1 mole of NaHCO3 produces 1 moleof CO2 gas for 0.9266 mol      it produces the same number of molesof CO2 .      The volume occupied by theCO2 gas is :            V= nRT / P                = 0.9266 mol x 0.0821 atm-L / mol-K x 295 K / 1atm                 = 22.34 L NaHCO3 (s) + CH3COOH (aq) =>CO2 (g) + H2O (l) +NaC2H3O2 (aq)         number of molesof NaHCO3 = 77.5 g / 84 g /mol                                                          = 0.9266 mol       From the balanced equatin it isclear that 1 mole of NaHCO3 produces 1 moleof CO2 gas for 0.9266 mol      it produces the same number of molesof CO2 .      The volume occupied by theCO2 gas is :            V= nRT / P                = 0.9266 mol x 0.0821 atm-L / mol-K x 295 K / 1atm                 = 22.34 L

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