Question

Aqueous hydrobromic acid (HBr) will react with solid sodium hydroxide (NaOH) to produce aqueous sodium bromide (NaBr) and liquid water (H2O). Suppose 5.7 g of hydrobromic acid is mixed with 0.980 g of sodium hydroxide. Calculate the maximum mass of water that could be produced by the chemical reaction. Round your answer to significant digits.

Answer 1

Answer:

The maximum mass of water that could be produced by the chemical reaction=0.441g

Explanation:

We are given that

Given mass of HBr=5.7 g

Given mass of sodium hydroxide=0.980 g

Molar mass of HBr=80.9 g/ Mole

Molar mass of NaOH=40 g/mole

Molar mass of H2O=18 g/mole

Reaction

$HBr+NaOH\rightarrow H_2O+NaBr$

Number of moles=$\frac{given\;mass}{molar\;mass}$

Using the formula

Number of moles of HBr=$\frac{5.7}{80.9}=0.0705 moles$

Number of moles of NaOH=$\frac{0.980}{40}=0.0245moles$

Hydrogen bromide is in a great excess and the amount of water produced.

Therefore,

Number of moles of water, n(H2O)=Number of moles of NaOH=0.0245moles

Now,

Mass of water=$n(H_2O)\times Molar\;mass\;of\;water$

Mass of water=$0.0245moles\times 18=0.441g$

Hence, the maximum mass of water that could be produced by the chemical reaction=0.441g