Aqueous hydrobromic acid reacts with solid sodium hydroxide to produce aqueous sodium bromide and liquid water . If of water is produced from the reaction of of hydrobromic acid and of sodium hydroxide, calculate the percent yield of water. Round your answer to significant figures.

See explanation below

Explanation:

You are missing certain data. I found one, in the attached picture. All you have to do is replace your data in this procedure to get an accurate answer.

According to the picture attached, we have 2.4 g of HBr and 1.9 g of NaOH. This mix produces 0.411 g of water. To get the %yield of water, we first need to see how many theorical grams of water are formed. With this we can calculate the %yield.

The overall reaction is:

HBr + NaOH --------> NaBr + H₂O

We have a mole ratio of 1:1 so, let's see which is the limiting reactant between the acid and the base, using the reported molar mass for each (MM HBr = 80.9119 g/mol; MM NaOH = 39.997 g/mol)

moles HBr = 2.4 / 80.9119 = 0.02966 moles

moles NaOH = 1.9 / 39.997 = 0.0475 moles

So, we can clearly see that the moles of NaOH are in excess, so the HBr is the limiting reactant.

As we have a mole ratio of 1:1 with each compound here, and that the HBr is the limiting reactant we can say that:

moles HBr = moles H₂O = 0.02966 moles of water

Now, using the molar mass of water, let's calculate the theorical yield of water: (MM water = 18.01528 g/mol

m H₂O = 0.02966 * 18.01528 = 0.534 g

We have the theorical yield, let's calculate the % yield:

% = 0.411 / 0.534 * 100

% = 76.92%