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.

Question

Madalyn Potter

Chemistry

24 December, 10:56

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)

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Hassan Shaw · Answer 1

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

Sergio Small · Answer 2

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