Question

Suppose you have 20.00 mL of a .100 M AgNO3 (aq) solution. you add 25 mL of .0800 M NaBr (aq) solution. How many grams of solid AgBr can you isolate?

Answer 1

moles of AgNO3 = 20 x 0.1 / 1000 = 2 x 10^-3

moles of NaBr = 25 x 0.08 / 1000 = 2 x 10^-3

AgNO3 + NaBr ----------------> AgBr + NaNO3

1               1                              1

so moles of AgBr formed = 2 x 10^-3

mass of AgBr =moles x molar mass

                       = 2 x 10^-3 x 187.77

                      = 0.376 g

mass of solid AgBr = 0.376 g