Here's how you do the calculations.
PROBLEM:
A student added 50.00 mL of 0.1000 mol/L #"HCl"# to 25.00 mL of a commercial ammonia-based cleaner.
It took 21.50 mL of 0.1000 mol/L #"NaOH"# to neutralize the excess #"HCl"#.
What was the concentration of ammonia in the cleaner?
Solution:
#"Part 1. HCl calculations"#
(a) Calculate the moles of #"HCl"# added to the cleaning solution
#"Moles of HCl" = 0.05000 cancel("L HCl") × "0.1000 mol HCl"/(1 cancel("L HCl")) = "0.005 000 mol HCl"#
(b) Calculate the moles of #"NaOH"# used
#"Moles of NaOH" = 0.021 50 cancel("L NaOH") × "0.1000 mol NaOH"/(1 cancel("L NaOH")) = "0.002 150 mol NaOH"#
(c) Calculate the moles of excess #"HCl"#
#"HCl + NaOH" → "NaCl + H"_2"O"#
#"Moles of HCl" = 0.002 150 cancel("mol NaOH") × "1 mol HCl"/(1 cancel("mol NaOH")) = "0.002 150 mol HCl"#
(d) Calculate the moles of #"HCl"# that reacted with the #"NH"_3#
#"Moles of HCl reacted = 0.005 000 mol – 0.002 150 mol = 0.002 850 mol"#
#"Part 2. NH"_3 color(white)(l)"calculations"#
(a) Calculate the moles of #"NH"_3#
#"NH"_3 + "HCl" → "NH"_4"Cl"#
#"Moles of NH"_3 = 0.002 850 cancel("mol HCl") × ("1 mol NH"_3)/(1 cancel("mol HCl")) = "0.002 850 mol NH"_3#
(b) Calculate the molarity of the #"NH"_3#
#"Molarity" = "0.002 850 mol"/"0.02500 L" = "0.1140 mol/L"#
