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20 March, 06:01

If 10.0 grams of NaHCO3 is added to 10.0 g of HCl, determine the efficiency of baking soda as an antacid if 6.73 g of NaCl was produced from the reaction. (In other words, calculate the percent yield of sodium chloride produced).

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  1. 20 March, 08:08
    0
    The percent yield of this reaction is 96.8 %

    Explanation:

    Step 1: Data given

    Mass of NaHCO3 = 10.0 grams

    Mass of HCl = 10.0 grams

    MAss of NaCl produced = 6.73 grams

    Molar mass of NaHCO3 = 84.0 g/mol

    Molar mass HCl = 36.46 g/mol

    Molar mass NaCl = 58.44 g/mol

    Step 2: The balanced equation

    NaHCO3 (aq) + HCl (aq) → NaCl (aq) + CO2 (g) + H2O (l)

    Step 3: Calculate moles

    Moles = mass / molar mass

    Moles NaHCO3 = 10.0 grams / 84.0 g/mol

    Moles NaHCO3 = 0.119 moles

    Moles HCl = 10.0 / 36.46 g/mol

    Moles HCl = 0.274 moles

    Step4: Calculate limiting reactant

    For 1 mol NaHCO3 we need 1 mol HCl to produce 1 mol NaCl, 1 mol CO2 and 1 mol H2O

    NaHCO3 is the limiting reactant. It will completely be consumed (0.119 moles). HCl is in excess. There will react 0.119 moles. There will remain 0.274 - 0.119 = 0.155 moles

    Step 5: Calculate moles NaCl

    For 1 mol NaHCO3 we need 1 mol HCl to produce 1 mol NaCl, 1 mol CO2 and 1 mol H2O

    For 0.119 moles NaHCO3 we'll have 0.119 moles NaCl

    Step 6: Calculate mass NaCl

    Mass NaCl = moles NaCl * molar mass NaCl

    Mass NaCl = 0.119 moles NaCl * 58.44 g/mol

    Mass NaCl = 6.95 grams

    Step 7: Calculate the percent yield

    Percent yield = (actual mass / theoretical mass) * 100%

    Percent yield = (6.73 grams / 6.95 grams) * 100%

    PErcent yield = 96.8 %

    The percent yield of this reaction is 96.8 %
  2. 20 March, 09:55
    0
    percentage yield of NaCl = 96.64%

    Explanation:

    The reaction was between NaHCO3 and HCl. The chemical equation can be represented below:

    NaHCO3 + HCl → NaCl + H2O + CO2. The balance equation is

    NaHCO3 + HCl → NaCl + H2O + CO2

    The question ask us to calculate the percentage yield of NaCl.

    The efficiency of NaHCO3 as an antacid, the limiting reactant is NaHCO3

    as

    1 mole of NaHCO3 produces 1 mole of NaCl

    Therefore,

    molar mass of NaHCO3 = 23 + 1 + 12 + 48 = 84 g

    molar mass of NaCl = 23 + 35.5 = 58.5 g

    1 mole of NaHCO3 = 84 g

    1 mole of NaCl = 58.5 g

    since 84 g of NaHCO3 produces 58.5 g of NaCl

    10 g of NaHCO3 will produce? grams of NaCl

    cross multiply

    Theoretical yield of NaCl = (10 * 58.5) / 84

    Theoretical yield of NaCl = 585/84

    Theoretical yield of NaCl = 6.9642857143 g

    percentage yield of NaCl = actual yield/theoretical yield * 100

    percentage yield of NaCl = 6.73/6.9642857143 * 100

    percentage yield of NaCl = 673/6.9642857143

    percentage yield of NaCl = 96.635897436%

    percentage yield of NaCl = 96.64%
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