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

Question

Skyler Randolph

Chemistry

23 August, 18:23

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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Answers (2)

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Phillip James · Answer 1

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.

Aidan Wells · Answer 2

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