Question

Solutions of sodium carbonate and silver nitrate react to form solid silver carbonate and a solution of sodium nitrate. A solution containing 8.75 g of sodium carbonate is mixed with one containing 6.00 g of silver nitrate. After the reaction is complete, the solutions are evaporated to dryness, leaving a mixture of salts. How many grams of each of the following compounds are present after the reaction is complete?

a) sodium carbonate

b) silver nitrate

c) silver carbonate

d) sodium nitrate

Answer 1

we know that

moles = mass / molar mass

so

moles of Na2C03 taken = 8.75 / 106 = 0.082547

moles of AgN03 taken = 6 / 170 = 0.0356

now

the reaction is

Na2C03 + 2AgN03 ---> Ag2C03 + 2NaN03

we can see that

moles of AgN03 required = 2 x moles of Na2C03

so

moles of AgN03 required = 2 x 0.082547 = 0.165

but

only 0.0356 moles of AgN03 is present

so

AgN03 is the limiting reagent

from the reaction we can see that

moles of Na2C03 reacted = 0.5 x moles of AgN03

moles of Na2C03 reacted = 0.5 x 0.0356 = 0.0178

so

moles of Na2C03 unreacted = 0.082547 - 0.0178 = 0.064747

now

mass = moles x molar mass

so

mass of Na2C03 left = 0.064747 x 106 = 6.863

now

moles of Ag2C03 formed = moles of Na2C03 reacted = 0.0178

mass of Ag2C03 formed = 0.0178 x 275.75 = 4.91 grams

moles of NaN03 formed = moles of AgN03 reacted = 0.0356

mass of AgN03 formed = 0.0356 x 85 = 3.026 grams

so

a) Na2C03 ---> 6.863 grams

b) AgN03 --> 0

c) Ag2C03 ---> 4.91 grams

d) NaN03 ---> 3.026 grams