Answer to Question #249722 in General Chemistry for hhahaa

Question

Solutions of sodium carbonate and silver nitrate react to form solid silver carbonate and a solution of sodium nitrate. A solution containing 3.20 g

g of sodium carbonate is mixed with one containing 4.57 g

g of silver nitrate. How many grams of silver carbonate are present after the reaction is complete? How many grams of sodium nitrate are present after the reaction is complete?

Answer 1

Na2CO3 + 2AgNO3 → Ag2CO3 + 2NaNO3

n(Na2CO3) = m(Na2CO3) / M(Na2CO3) = 3.20 g / 106 g/mol = 0.0301 mol

n(AgNO3) = m(AgNO3) / M(AgNO3) = 4.57 g / 170 g/mol = 0.0268 mol

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

For Ag2CO3:

0.0268 mol AgNO3 x (1 mol Ag2CO3 / 2 mol AgNO3 ) = 0.0134 mol Ag2CO3

0.0134 mol Ag2CO3 x 275.7 g/mol = 3.69 g Ag2CO3

For NaNO3

0.0268 mol AgNO3 x (2 mol NaNO3 / 2 mol AgNO3 ) = 0.0268 mol NaNO3

0.0268 mol NaNO3 x 85 g/mol = 2.278 g NaNO3

Ansver:

1) 3.69 g Ag2CO3

2) 2.278 g NaNO3