Answer to Question #165152 in General Chemistry for Claire

Question

What mass of Ag2CO3(275.7 g/mol) is formed when 25.0 ml of 0.200 M. AgNO3 are mixed with 50.0 ml of 0.0800 M Na2CO3?

Answer 1

2AgNO3 + Na2CO3 ---> Ag2CO3 + 2NaNO3

Moles of AgNO3= C x V

= 0.200 x 25.0/1000 = 5 x 10^-3mol

Mole of Na2CO3= 0.0800 x 50.0/1000 = 4 x 10^-3mol

Mass of AgNO3= mole x molar mass

= 5x10^-3 x 170= 0.85g

Mass of Na2CO3= 4 x 10^-3 x 106

= 0.424g

Now let's find the limiting reagent that determines the mass of AgCO3 formed. From the equation of reaction.

1mol of Na2CO3 reacts with 2mol of AgNO3

106g of Na2CO3 reacts with 340g of AgNO3

0.424g of Na2CO3 should react with 340/160 x 0.424 = 0.901g

Since only 0.85g of AgNO3 is available, it is the limiting reagent.

340g of AgNO3 produces 276g of Ag2CO3

0.85g of AgNO3 will produce 276/340 x 0.85 = 0.69g

Therefore, the mass of Ag2CO3 produced is 0.69g