Question

Solutions of sodium carbonate and silver nitrate react to form solid silver carbonate and a solution of sodium nitrate. A solution containing 3.50 g of sodium carbonate is mixed with one containing 5.00 of silver nitrate.

Part A)
How many grams of sodium carbonate are present after the reaction is complete?

Part B)
How many grams of silver nitrate are present after the reaction is complete?

Part C)
How many grams of silver carbonate are present after the reaction is complete?

Part D)
How many grams of sodium nitrate are present after the reaction is complete?

Answer 1

Ok so first you want to write out the equation so you know where to start from.

Na2CO3 + 2 AgNO3 Ag2CO3 + 2 NaNO3

Next it's important to recognize that this is a limiting reagent problem where one of the reagents will be completely used up. To find this you must use moles instead of grams.

moles Na2CO3 = 3.50 g / 106 g/mol = 0.0330 mol

moles AgNO3 = 5.00 g / 170 g/mol = 0.0294 mol

Determine the limiting reagent: 0.0330 mol Na2CO3 * (2 mol AgNO3 / 1 mol Na2CO3) = 0.0660 mol AgNO3 is required to react with Na2CO3 but there is not enough AgNO3 so it is limiting.

Using stoichiometry:

0.0294 mol AgNO3 * ( 1 mol Na2CO3 / 2 mol AgNO3) = 0.0147 mol Na2CO3 will react

0.0330 mol Na2CO3 - 0.0147 mol Na2CO3 = 0.0183 mol Na2CO3 remains or

0.0183 mol Na2CO3 = 1.94 g Na2CO3

Use moles AgNO3 to determine moles and mass of products in reaction by stoichiometry.

For Ag2CO3:

0.0294 mol AgNO3 * (1 mol Ag2CO3 / 2 mol  AgNO3 ) = 0.0147 mol Ag2CO3

0.0147 mol Ag2CO3 * 275.7 g/mol = 4.05 g Ag2CO3

For NaNO3

0.0294 mol AgNO3 * (2 mol NaNO3 / 2 mol  AgNO3 ) = 0.0294 mol NaNO3

0.0294 mol NaNO3 * 85 g/mol = 2.50 g NaNO3

So to answer your problem. Hope this helps!

A) 1.94 g Na2CO3

B) 0.00 g AgNO3

C) 4.05 g Ag2CO3

D) 2.50 g NaNO3