Aqueous hydrochloric acid HCl will react with solid sodium hydroxide NaOH to produce aqueous sodium chloride NaCl and liquid water H2O. Suppose 0.365 g of hydrochloric acid is mixed with 0.18 g of sodium hydroxide. Calculate the maximum mass of sodium chloride that could be produced by the chemical reaction. Round your answer to 2 significant digits.

0.26g of NaCl is the maximum mass that could be produced

Explanation:

Based on the reaction:

HCl + NaOH → NaCl + H₂O

Where 1 mol of HCl reacts per mol of NaOH to produce 1 mol of NaCl

To solve this question we need to find limiting reactant. The moles of limiting reactant = Moles of NaCl produced:

Moles HCl -Molar mass: 36.46g/mol-:

0.365g HCl * (1mol / 36.46g) = 0.010 moles HCl

Moles NaOH -Molar mass: 40g/mol-:

0.18g NaOH * (1mol / 40g) = 0.0045 moles NaOH

As the reaction is 1:1 and moles NaOH < moles HCl, limiting reactant is NaOH and maximum moles produced of NaCl are 0.0045 moles.

The mass of NaCl is:

Mass NaCl -Molar mass: 58.44g/mol-:

0.0045 moles * (58.44g/mol) =

the correct answer is:

a. formation of precipitation

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