Question

A solution containing 5.50 g of sodium carbonate is mixed with one containing 6.50 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?

Answer 1

we know that

moles = mass / molar mass

so

moles of Na2C03 taken = 5.5 / 106 = 0.052

moles of AgN03 taken = 6.5 / 170 = 0.03823

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 0.052 = 0.104

but

only 0.03823 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 0.03823 = 0.019115

so

moles of Na2C03 unreacted = 0.052 - 0.019115 = 0.032885

now

mass = moles x molar mass

so

mass of Na2C03 left = 0.032885 x 106 = 3.486

now

moles of Ag2C03 formed = moles of Na2C03 reacted = 0.019115

mass of Ag2C03 formed = 0.019115 x 275.75 = 5.271 grams

moles of NaN03 formed = moles of AgN03 reacted = 0.03823

mass of NaN03 formed = 0.03823 x 85 = 3.25 grams

so

a) Na2C03 ---> 3.486 grams

b) AgN03 --> 0

c) Ag2C03 ---> 5.271 grams

d) NaN03 ---> 3.25 grams