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20 February, 01:37

In a certain experiment 1.00 g of sodium bicarbonate and 1.00 g of citric acid are allowed to react. (a) which is the limiting reactant? (b) how many grams of carbon dioxide form? (c) how many

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  1. 20 February, 01:48
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    The complete balanced chemical reaction for this is:

    C6H8O7 + 3 NaHCO3 - - - > 3 H2O + 3 CO2 + Na3C6H5O7

    First calculate for the number of moles of each reactant.

    moles C6H8O7 = 1 g / (192.124 g / mol) = 5.2 * 10^-3 mol

    moles NaHCO3 = 1 g / (84.01 g / mol) = 11.9 * 10^-3 mol

    A. The ratio of the reactant from the chemical reaction is 3NaHCO3:1C6H8O7, while the given chemicals are in the ratio of:

    11.9 * 10^-3NaHCO3: 5.2 * 10^-3 C6H8O7 = 2.29NaHCO3:1C6H8O7

    Therefore this means that there is less amount of NaHCO3 supplied than what is required therefore the limiting reactant is:

    NaHCO3

    B. We calculate based on the limiting reactant.

    mass CO2 = 11.9 * 10^-3 mol NaHCO3 (3 mol CO2/1 mol NaHCO3) (44.01 g/mol)

    mass CO2 = 1.57 g

    C. I believe what is asked here is the amount of excess reactant which remains. The excess reactant is C6H8O7.

    mass C6H8O7 left = [5.2 * 10^-3 mol - (11.9 * 10^-3 mol NaHCO3 (1 mol C6H8O7 / 3 mol NaHCO3)) ] * (192.124 g / mol)

    mass C6H8O7 left = 0.237 g
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