Question

Aqueous hydrochloric acid (HCI) will react with solid sodium hydroxide (NaOH) to produce aqueous sodium chloride (NaCl) and liquid water (HO) Suppose 6.93 g of hydrochloric acid is mixed with 13. g of sodium hydroxide. Calculate the maximum mass of sodium chloride that could be produced by the chemical reaction. Be sure your answer has the correct number of significant digits. 2 Explanation Chec

Answer 1

The balanced reaction between HCl and NaOH is as follows:

HCl + NaOH $\rightarrow$ NaCl + H₂O

Mass of HCl = 6.93 g

Mass of NaOH = 13. g

Determine the number of moles of HCl using mass and molar mass as follows:

The molar mass of HCl = 36.46 g/mol

= 6.93 g HCl x ( 1 mol HCl / 36.46 g HCl)

= 0.1901 mol HCl

Similarly,

Determine the number of moles of NaOH using mass and molar mass as follows:

The molar mass of NaOH = 39.997 g/mol

= 13. g NaOH x ( 1 mol NaOH / 39.997 g NaOH)

= 0.3250 mol NaOH

Determine the limiting reactant as follows:

Use the moles of NaOH and mole ratio from the balanced chemical reaction. Determine the number of moles of HCl required to react completely with 0.3250 mol NaOH as follows:

= 0.3250 mol NaOH x ( 1 mol HCl/ 1 mol NaOH)

= 0.3250 mol HCl is required to react completely with 0.3250 mol NaOH.

Similarly, Use the moles of HCl and mole ratio from the balanced chemical reaction. Determine the number of moles of NaOH required to react completely with 0.1901 mol HCl as follows:

= 0.1901 mol HCl x ( 1 mol NaOH / 1 mol HCl)

= 0.1901 mol NaOH is required to react completely with 0.1901 mol HCl.

Now, moles of HCl required to react completely with the given moles of NaOH(0.3250 mol) is more than the given moles of HCl (0.1901 mol).

Thus, HCl is the limiting reactant.

Use the moles of limiting reactant and mole ratio from the balanced chemical reaction. Determine the number of moles of NaCl as follows:

= 0.1901 mol HCl x ( 1 mol NaCl / 1 mol HCl)

= 0.1901 mol NaCl will be produced.

Convert 0.1901 mol NaCl to grams as follows:

The molar mass of NaCl = 58.44 g/mol

= 0.1901 mol NaCl x ( 58.44 g NaCl / 1 mol NaCl)

= 11.g NaCl will be produced.

Thus, the mass of NaCl that could be produced by the chemical reaction is 11.g. [2 S.F]