Question

Aqueous hydrobromic acid (HBr) will react with solid sodium hydroxide (NaOH) to produce aqueous sodium bromide (NaBr) and liquidwater (H₂O). Suppose 0.81 g of hydrobromic acid is mixed with 0.568 g of sodium hydroxide. Calculate the maximum mass of sodiumbromide that could be produced by the chemical reaction. Round your answer to 2 significant digits.

Answer 1

Explanation:

Aqueous hydrobromic acid (HBr) will react with solid sodium hydroxide (NaOH) to produce aqueous sodium bromide (NaBr) and liquid water (H₂O). They will react according to the following equation.

HBr + NaOH ---> NaBr + H₂O

0.81 g of HBr are mixed with 0.568 g of NaOH. We have to find the mass of NaBr that can be produced. To do that we have to find which of the reactants is limiting the reaction. First, we will convert their grams into moles using their molar masses.

molar mass of HBr = 80.91 g/mol

molar mass of NaOH = 40.00 g/mol

mass of HBr = 0.81 g

mass of NaOH = 0.568 g

moles of HBr = 0.81 g * 1 mol/(80.91 g)

moles of HBr = 0.0100 moles

moles of NaOH = 0.568 g * 1 mol/(40.00 g)

moles of NaOH = 0.0142 moles

HBr + NaOH ---> NaBr + H₂O

Now if we take a quick look at the coefficients of the reaction we will see that 1 mol of HBr will react with 1 mol of NaOH since both coefficients are 1. Then their molar ratio is 1 : 1. That also means that 0.0100 moles of HBr will only react with 0.0100 moles of NaOH, and we have mixed 0.0142 moles of it. So, NaOH is in excess and HBr is the limiting reagent.

1 mol of HBr : 1 mol of NaOH molar ratio

moles of NaOH = 0.0100 moles of HBr * 1 mol of NaOH/(1 mol of HBr)

moles of NaOH = 0.0100 moles < 0.0142 moles ----> NaOH is in excess

And now that we know that HBr is the limiting reagent we can find the number of moles of NaBr that will be produced by 0.0100 moles of HBr. And finally convert those moles into grams using the molar mass.

1 mol of HBr : 1 mol of NaBr molar ratio

moles of NaBr = 0.0100 moles of HBr * 1 mol of NaBr/(1 mol of HBr)

moles of NaBr = 0.0100 moles

molar mass of NaBr = 102.89 g/mol

mass of NaBr = 0.0100 moles * 102.89 g/mol

mass of NaBr = 1.0289 g

mass of NaBr = 1.0 g

Answer: the maximum mass of sodium bromide that could be produced is 1.0 g.