Question

A 5.95-g sample of \(\mathrm{AgNO}_{3}\) is reacted with \(\mathrm{BaCl}_{2}\) according to the equation

$$ 2 \mathrm{AgNO}_{3}(\mathrm{aq})+\mathrm{BaCl}_{2}(\mathrm{aq})->2 \mathrm{AgCl}(\mathrm{s})+\mathrm{Ba}\left(\mathrm{NO}_{3}\right)_{2}(\mathrm{aq}) $$

to give \(4.00 \mathrm{~g}\) of \(\mathrm{AgCl}\). What is the percent yield of \(\mathrm{AgCl} ?\)

a) \(21.4 \%\)

b) \(39.8 \%\)

c) \(67.2 \%\)

d) \(79.7 \%\)

e) \(93.5 \%\)

Answer 1

molar mass of AgNO3 = 169.9 g/mol
number of mol of AgNO3 = (given mass)/(molar mass)
= 5.95/169.9
= 0.0350 mol

since the coefficient of AgNO3 and AgCl is the same in the balanced reaction
so,
number of mol of AgCl = number of mol of AgNO3
= 0.0350 mol

molar mass of AgCl = 143.3 g/mol
mass of AgCl formed = (molar mass of AgCl)*(number of mol of AgCl)
= 143.3*0.0350
= 5.016 g

%yield = {(actual yield)/(theoretical yield)}*100
= (4.00/5.016) *100
= 79.7 %

Answer: option d