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20 December, 13:31

Aqueous hydrobromic acid will react with solid sodium hydroxide to produce aqueous sodium bromide and liquid water. Suppose 1.6 g of hydrobromic acid is mixed with 1.15 g of sodium hydroxide. Calculate the maximum mass of water that could be produced by the chemical reaction. Round your answer to significant digits.

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Answers (2)
  1. 20 December, 16:11
    0
    0.35 g of H₂O

    Explanation:

    Firstly we have to make the reaction's equation.

    In this case, the reactants are HBr and NaOH and the products are water and NaBr. This is a neutralization reaction.

    We calculate the moles of each reactant:

    1.6 g / 80.9 g/mol = 0.0198 moles of HBr

    1.15 g / 39 g/mol = 0.0295 moles of NaOH

    Reaction: NaOH + HBr → NaBr + H₂O

    Ratio is 1:1, so the limiting reactant is the hydrobromic acid.

    For 0.0295 moles of hydroxide we need the same amount of acid, but we do not have enough HBr, just 0.0198 moles.

    Ratio is 1:1, again. 1 mol of acid can produce 1 mol of water.

    Therefore, 0.0198 moles of HBr will produce 0.0198 moles of water.

    We convert the moles to mass → 0.0198 mol. 18 g/mol = 0.35 g of H₂O
  2. 20 December, 16:33
    0
    0.36 grams of H2O will be produced

    Explanation:

    Step 1: data given

    Mass of hydrobromic acid (HBr) = 1.6 grams

    Molar mass HBr = 80.91 g/mol

    Mass of sodium hydroxide (NaOH) = 1.15 grams

    Molar mass of NaOH = 40.0 g/mol

    Step 2: The balanced equation

    HBr + NaOH → NaBr + H2O

    Step 3: Calculate moles HBr

    Moles HBr = mass HBr / molar mass HBr

    Moles HBr = 1.6 grams / 80.91 g/mol

    Moles HBr = 0.0198 moles

    Step 4: Calculate moles NaOH

    Moles NaOH = 1.15 grams / 40.0 g/mol

    Moles NaOH = 0.0288 moles

    Step 5: Calculate limiting reactant

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

    HBr is the limiting reactant. It will completely be consumed (0.0198 moles). NaOH is in excess. There will react 0.019 moles. There will remain 0.0288 - 0.0198 = 0.009 moles

    Step 6: Calculate moles H2O

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

    For 0.0198 moles HBr we'll have 0.0198 moles of H2O

    Step 7: Calculate mass H2O

    Mass H2O = 0.0198 moles * 18.02 g/mol

    Mass H2O = 0.36 grams

    0.36 grams of H2O will be produced
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