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25 February, 01:43

Aqueous hydrobromic acid will react with solid sodium hydroxide to produce aqueous sodium bromide and liquid water. Suppose 68.0 g of hydrobromic acid is mixed with 25. g of sodium hydroxide. Calculate the maximum mass of water that could be produced by the chemical reaction. Be sure your answer has the correct number of significant digits

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  1. 25 February, 02:16
    0
    11.25 g of water, is the maximum mass that can be produced by the reaction

    Explanation:

    We propose the reaction, where aqueous hydrobromic acid reacts with solid sodium hydroxide to produce aqueous sodium bromide and liquid water

    The equation is: HBr (aq) + NaOH (s) → NaBr (aq) + H₂O (l)

    Ratio is 1:1.

    We convert the mass of the reactants to moles, in order to find out the limiting reagent

    68 g / 80.90 g/mol = 0.840 moles of HBr

    25 g / 40 g/mol = 0.625 moles of NaOH

    Limiting reagent is the NaOH. For 0.840 moles of HBr, we need 0.840 moles of NaOH, but we only have 0.625.

    To calculate the mass of water that could be produced, the ratio is 1:1

    Then, 0.625 moles of NaOH will produce 0.625 moles of water.

    We convert the moles to mass, to determine the maximum mass that can be produced → 0.625 mol. 18 g / 1mol = 11.25 g
  2. 25 February, 04:00
    0
    11.3 grams of water will be produced.

    Explanation:

    Step 1: Data given

    Mass of hydrobromic acid (HBr) = 68.0 grams

    Molar mass of HBr = 80.91 g/mol

    Mass of sodium hydroxide (NaOH) = 25.0 grams

    Molar mass of NaOH = 40.0 g/mol

    Step 2: The balanced equation

    HBr + NaOH → NaBr + H2O

    Step 3: Calculate moles

    Moles = mass / molar mass

    Moles HBr = 68.0 grams / 80.91 g/mol

    Moles HBr = 0.840moles

    Moles NaOH = 25.0 grams / 40.0 g/mol

    Moles NaOH = 0.625 moles

    Step 4: Calculate the limiting reactant

    NaOH is the limiting reactant. It will completely be consumed (0.625 moles).

    HBr is in excess. There will react 0.625 moles. There will remain 0.840 - 0.625 = 0.215 moles NaOH

    Step 5: Calculate moles H2O

    For 1 mol HBr we need 1 mol NaOH to produce 1 mol NaBr and 1 mol H2O

    For 0.625 moles NaOH we'll have 0.625 moles H2O

    Step 6: Calculate mass of H2O

    Mass of H2O = moles H2O * molar mass H2O

    Mass H2O = 0.625 moles * 18.02 g/mol

    Mass H2O = 11.3 grams H2O

    11.3 grams of water will be produced.
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