Aqueous hydrochloric acid will react with solid sodium hydroxide to produce aqueous sodium chloride and liquid water . Suppose 1.82 g of hydrochloric acid is mixed with 0.88 g of sodium hydroxide. Calculate the maximum mass of sodium chloride that could be produced by the chemical reaction. Round your answer to significant digits.

1.29 g of NaCl.

Explanation:

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

HCl(aq) + NaOH(s) —> NaCl(aq) + H2O(l)

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

This is illustrated below:

Molar mass of HCl = 1 + 35.5 = 36.5 g/mol

Mass of HCl from the balanced equation = 1 x 36.5 = 36.5 g

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

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

Molar mass of NaCl = 23 + 35.5 = 58.5 g/mol

Mass of NaCl from the balanced equation = 1 x 58.5 = 58.5 g

Summary:

From the balanced equation above,

36.5 g of HCl reacted with 40 g of NaOH to produce 58.5 g of NaCl.

Next, we shall determine the limiting reactant.

The limiting reactant can be obtained as follow:

From the balanced equation above,

36.5 g of HCl reacted with 40 g of NaOH.

Therefore, 1.82 g of HCl will react with = (1.82 x 40)/36.5 = 1.99 g of NaOH

From the calculations made above,

We can see that it will take a higher mass of NaOH i.e 1.99 g than what was given i.e 0.88 g to react completely with 1.82 g of HCl.

Therefore, NaOH is the limiting reactant and HCl is the excess reactant.

Finally, we shall determine the maximum mass of NaCl produced from the reaction.

In this case the limiting reactant will be used as it will produce the maximum mass of the products since all of it were consumed in the reaction.

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

From the balanced equation above,

40 g of NaOH reacted to produce 58.5 g of NaCl.

Therefore, 0.88 g of NaOH will react to produce = (0.88 x 58.5)/40 = 1.29 g of NaCl.

Therefore, 1.29 g of NaCl were produced from the reaction.

2b2t

explanation:

2b2t