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Aqueous hydrobromic acid HBr will react with solid sodium hydroxide NaOH to produce aqueous sodium bromide NaBr and liquid water H2O. Suppose 55.8 g of hydrobromic acid is mixed with 17. g of sodium hydroxide. Calculate the minimum mass of hydrobromic acid that could be left over by the chemical reaction. Round your answer to 2 significant digits.

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Answer:

21.4g of HBr is the minimum mass that could be left over.

Explanation:

Based on the reaction:

HBr + NaOH → NaBr + H2O

1 mole of HBr reacts per mole of NaOH

To solve this question we need to find the moles of both reactants. If moles NaOH > moles HBr, the difference in moles represents the minimum moles of HBr that could be left over because this reaction is 1:1. Using the molar mass we can find the minimum mass of HBr that could be left over, as follows:

Moles NaOH -40.0g/mol-

17g * (1mol/40.0g) = 0.425 moles NaOH

Moles HBr -Molar mass: 80.91g/mol-

55.8g * (1mol/80.91g) = 0.690 moles HBr

The difference in moles is:

0.690 moles - 0.425 moles =

0.265 moles of HBr could be left over

The mass is:

0.265 moles * (80.91g/mol) =

21.4g of HBr is the minimum mass that could be left over.

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