Aqueous hydrobromic acid will react with solid sodium hydroxide to produce aqueous sodium bromide and liquid water . Suppose 34. g of hydrobromic acid is mixed with 11.4 g of sodium hydroxide. Calculate the maximum mass of water that could be produced by the chemical reaction

5.13g of H2O.

Explanation:

We'll begin by writing the balanced equation for the reaction. This is given below:

HBr(aq) + NaOH(aq) —> NaBr(aq) + H2O(l)

Next, we shall determine the masses of HBr and NaOH that reacted and the mass of H2O produced from the balanced equation.

This is illustrated below:

Molar mass of HBr = 1 + 80 = = 81g/mol

Mass of HBr from the balanced equation = 1 x 81 = 81g

Molar mass of NaOH = 23 + 16 + 1 = 40g/mol

Mass of NaOH from the balanced equation = 1 x 40 = 40g

Molar mass of H2O = (2x1) + 16 = 18g/mol

Mass of H2O from the balanced equation = 1 x 18 = 18g

Thus, from the balanced equation above,

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

Next, we shall determine the limiting reactant. This is illustrated below:

From the balanced equation above,

81g of HBr reacted with 40g of NaOH.

Therefore, 34g of HBr will react with = (34 x 40)/81 = 16.79g of NaOH.

From the calculations made above, we can see that it will take a higher mass (i.e 16.79g) of NaOH than what was given (i.e 11.4g) to react completely with 34g of HBr.

Therefore, NaOH is the limiting reactant and HBr is the excess react.

Finally, we can determine the maximum mass of H2O produced as shown below.

In this case the limiting reactant will used because it will produce the maximum mass of H2O since all of it is consumed in the reaction.

The limiting reactant is NaOH and the maximum mass of H2O produced can be obtained as follow:

From the balanced equation above,

40g of NaOH reacted to produce 18g of H2O.

Therefore, 11.4g of NaOH will react to produce = (11.4 x 18)/40 = 5.13g of H2O.

Therefore, the maximum mass of H2O produced from the reaction is 5.13g.

it's filtration, but this is an assumption so my apologies if i got this

c. 75.7

i think that is the right