20g of NaHCO3 Treated with 22g of HCl Producing 12g of the residue. What is the amount of co2 Formed ?

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  1. NaHCO3 + HCl ==> NaCl + H2O + CO2

    mols NaHCO3 = grams/molar mass = 20/84 = approx 0.24.

    mols CO2 formed if NaHCO3 is the limiting reagent (LR)is
    0.24 mols NaHCO3 x (1 mol CO2/1 mol NaHCO3) = 0.24 mols CO2.

    mols HCl = 22/35.5 = approx 0.62.
    mols CO2 formed if HCl is the LR is 0.6 x (1 mol CO2/1 mol HCl) = 0.6 mols CO2.

    In LR problems, the SMALLER amount is ALWAYS the correct answer; therefore, 0.24 mols CO2 will be formed.
    grams CO2 = 0.24 mol CO2 x molar mass CO2 = ? g is you want grams and not mols CO2.
    NOTE: I have estimated these numbers; you should redo all of the calculations to get better numbers.

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  2. I want it in grams

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  3. "I want" -- ??

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  4. Note that I showed how to do it in mols AND in grams.

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