Question

If 15.00 mL of a 1.00 M HCl solution is mixed with 25.00 mL of a 0.250 M NaOH solution in a process that produces 273 mg of NaCl, what is the percent yield?

Answer 1

we know that

moles = molarity x volume (L)

so

moles of HCl = 1 x 15 x 10-3 = 15 x 10-3

moles of NaoH = 0.25 x 25 x 10-3 = 6.25 x 10-3

now

the reaction is

HCl + NaOH ---> NaCl + H20

we can see that

moles of HCl reacted = moles of NaOH added

so

moles of HCl reacted = 6.25 x 10-3

but

15 x 10-3 moles of HCL is present

so

HCl is in excess and NaOH is the limiting reagent

now

consider the reaction

NaOH + HCl --> NaCl + H20

we can see that

moles of NaCl formed = moles of NaOH added = 6.25 x 10-3

now

we know that

mass = moles x molar mass

so

mass of NaCl produced = 6.25 x 10-3 x 58.44

mass of NaCl produced = 365.25 x 10-3 g

mass of NaCl produced = 365.25 mg

now

this is the theoretical mass of NaCl produced

but

given

actual mass of NaCl produced is 273 mg

so

% yield = actual x 100 / theoretical

so

% yield = 273 x 100 / 365.25

% yield = 74.74

so

percent yield is 74.74 %