Question

Problem Page Aqueous hydrobromic acid will react with solid sodium hydroxide to produce aqueous sodium bromide and liquid water . Suppose 80. g of hydrobromic acid is mixed with 55.0 g of sodium hydroxide. Calculate the minimum mass of hydrobromic acid that could be left over by the chemical reaction. Be sure your answer has the correct number of significant digits.

Answer 1

To produce 0.9888 moles of NaBr and 0.9888 moles of H2O, there will remain 0.3862 moles of NaOH. All the HBr will be consumed, no HBr will be left over by the chemical reaction

Step-by-step explanation:

Step 1: Data given

Mass of hydrobromic acid = 80.00 grams

Mass of sodium hydroxide = 55.00 grams

Molar mass HBr = 80.91 g/mol

Molar mass NaOH = 40 g/mol

Step 2: The balanced equation

HBr(aq) + NaOH(s) → NaBr(aq) + H₂O(l)

Step 3: Calculate moles of HBr

Moles HBr = Mass HBr / molar mass HBr

Moles HBr = 80.00 grams / 80.91 g/mol

Moles HBr = 0.9888 moles

Step 4: Calculate moles NaOH

Moles NaOH = 55.00 grams / 40.00 g/mol

Moles NaOH = 1.375 moles

Step 5: Calculate limiting reactant

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

The limiting reactant is HBr. It will completely be consumed. (0.9888 moles). NaOH is in excess. There will react 0.9888 moles. There will remain 1.375 - 0.9888 = 0.3862 moles

This means to produce 0.9888 moles of NaBr and 0.9888 moles of H2O, there will remain 0.3862 moles of NaOH. All the HBr will be consumed, no HBr will be left over by the chemical reaction

Answer 2

No HBr will be left over by the chemical reaction.

Step-by-step explanation:

The reaction of hydrobromic acid with sodium hydroxide is:

HBr + NaOH → NaBr + H₂O

80,0g of HBr are:

80,0g×(1mol/80,91g) = 0,989 moles

55,0g of NaOH are:

55,0g×(1mol/40g) = 1,375 moles

That means excess moles of NaOH are:

1,375-0,989 = 0,286 moles of NaOH

Thus, HBr is the limiting reactant and will be used completely. As a result, no HBr will be left over by the chemical reaction.

I hope it helps!