Question

What mass of solid AgBr is produced when 100.0 ml of 0.150 M AgNO₃ is added to 20.0 ml of 1.00 M NaBr?

Answer 1

volume of AgNO3, V = 100.0 mL

= 0.1 L

we have below equation to be used:

number of mol in AgNO3,

n = Molarity * Volume

= 0.15*0.1

= 1.5*10^-2 mol

volume of NaBr, V = 20.0 mL

= 2*10^-2 L

we have below equation to be used:

number of mol in NaBr,

n = Molarity * Volume

= 1*0.02

= 2*10^-2 mol

we have the Balanced chemical equation as:

AgNO3 + NaBr ---> AgBr + NaNO3

1 mol of AgNO3 reacts with 1 mol of NaBr

for 1.5*10^-2 mol of AgNO3, 1.5*10^-2 mol of NaBr is required

But we have 2*10^-2 mol of NaBr

so, AgNO3 is limiting reagent

we will use AgNO3 in further calculation

Molar mass of AgBr = 1*MM(Ag) + 1*MM(Br)

= 1*107.9 + 1*79.9

= 187.8 g/mol

From balanced chemical reaction, we see that

when 1 mol of AgNO3 reacts, 1 mol of AgBr is formed

mol of AgBr formed = (1/1)* moles of AgNO3

= (1/1)*1.5*10^-2

= 1.5*10^-2 mol

we have below equation to be used:

mass of AgBr = number of mol * molar mass

= 1.5*10^-2*1.878*10^2

= 2.82 g

Answer: 2.82 g

Feel free to comment below if you have any doubts or if this answer do not work. I will edit it if you let me know