Aqueous hydrobromic acid will react with solid sodium hydroxide to produce aqueous sodium bromide and liquid water. Suppose 66. g of hydrobromic acid is mixed with 21.6 g of sodium hydroxide. Calculate the maximum mass of water that could be produced by the chemical reaction. Round your answer to significant digits.

9.72g

Explanation:

Step 1:

The balanced equation for the reaction. This is illustrated below:

HBr + NaOH —> NaBr + H2O

Step 2:

Determination of the masses of HBr and NaOH that reacted and the mass of H2O produced from the balanced equation.

From the equation above, 1 mole of HBr reacted with 1 mole NaOH to produce 1 mole of H2O

1 mole of HBr = 1 + 80 = 81g

1 mole of NaOH = 23 + 16 + 1 = 40g

1 mole of H2O = (2x1) + 16 = 18g

So, we can say that:

From the balanced equation above,

81g of HBr reacted with 40g of NaOH to produce 18g of H2O.

Step 3:

Determination of the limiting reactant. This is illustrated below:

From the balanced equation above,

81g of HBr reacted with 40g of NaOH.

Therefore, 66g of HBr will react with = (66 x 40)/81 = 32.6g of NaOH

We can see that the mass of NaOH (i.e 32.6g) obtained is far greater than was was given (21.6g). Therefore, NaOH is the limiting reactant.

Step 4:

Determination of the maximum mass of H2O produced from the reaction.

The limiting reactant is used to determine the maximum yield of the reaction.

The maximum mass of H2O produced is obtained as follow:

From the balanced equation above,

40g of NaOH produce 18g of H2O.

Therefore, 21.6g of NaOH will produce = (21.6 x 18)/40 = 9.72g of H2O.

The reaction produce 9.72g of water (H2O)

i think   the answer is c hope i'm right