NaOH(s) + HCl (aq) --> NaCl (aq) Determine the mass of salt formed when 60g of sodium hydroxide combined with 60g of hydrochloric acid. What is the mass of the reactant that remains when the reaction has stopped?

socratic

Answer:

Explanation:
Here’s our balanced equation:
#NaOH + HCl —> NaCl + H_2O#
From this, we know that, for every #1# mole of #NaOH# that reacts, #1# mole of #HCl# will react to form #1# mole of #NaCl#.
In other words, the mole ratio of #NaOH:HCl:NaCl# is #1:1:1#.

• In #"60 g"# of #NaOH#, there are #60/(22.99 xx 16.00 xx 1.008) = 1.50# moles of #NaOH#.
• In #"60 g"# of #HCl#, there are #60/(1.008 + 35.45) = 1.65# moles of #HCl#.

So, from this, we can see that #NaOH# is the limiting reactant.

When this reaction happens, #1.50# moles of #NaOH# will react with #1.50# moles of #HCl# (because their mole ratio is #1:1#) to form #1.50# moles of NaCl.

To find the mass of #1.50# moles of #NaCl#, we can multiply #1.50# moles by #NaCl#'s molar mass:

#1.50 xx (22.99 + 35.45) = "87.66 g"#

Since the question only had #1# significant figure, we’d need to round that to #"90 g"#.

After the #"90 g"# of #NaCl# is formed, there will still be #1.65-1.50 = 0.15# moles of the excess reactant, #HCl#, left.
To find this mass of #0.15# moles, we just need to multiply #0.15# by the molar mass of #HCl#:

#0.15 xx (1.008 + 35.45) = "5.47 g"#

Again, there's only one significant figure in the question, so we'd need to round #"5.47 g"# to #"6 g"#.