Insoluble silver carbonate, Ag2CO3(s), forms in the following balanced chemical reaction:

Question

2AgNO3(aq) + K2CO3(aq) = Ag2CO3(s) + 2KNO3(aq).

What mass of silver nitrate, 2AgNO3(aq), reacts with 25.0 g of potassium carbonate,K2CO3(aq),if there is at least 5.5 g of silver nitrate in excess

I have calculated the molar mass for the silver nitrate and for the potassium carbonate, and calculated the moles for the potassium carbonate as well. I am not sure what to do after this: the final answer is supposed to be 61.5 g.

Answer 1

U find moles of k2co3 then u do mike ratio between the two reactants once u do that u have the moles of agno3 u convert that to mass by m=nxM
Hope that helps!

Answer 2

Doing this method doesn’t equal out to 61.5

Answer 3

Gale buddy, how do you do it please

Answer 4

First, find the molar mass of AgNO3 and K2CO3 then use that molar mass (AgNO3 = 169.87 g/mol and K2CO3 = 138.205 g/mol) and their mass (AgNO3 = ? and K2CO3 = 25 g) to calculate the mol for both.

Remember Mol = Mass/Molar mass.

(Mole for K2CO3) = (25 g) / (138.205 g/mol) = (0.180884 mol)

Balanced chemical equation:

2AgNO3 + K2CO3 → Ag2CO3 + 2KNO3

Mole ratio: when there are 2 moles of AgNO3, there is 1 mole of K2CO3 so the mole ratio is 2:1 or 2/1

(? mol of AgNO3) / (0.180884 mol of K2CO3) = (2 mol of AgNO3) / (1 mol of K2CO3)

we cross multiply to get :
(2 mol of AgNO3) * (0.180884 mol of K2CO3) = (? mol of AgNO3) * (1 mol of K2CO3)

(2 mol of AgNO3) * (0.180884 mol of K2CO3) / (1 mol of K2CO3) = (? mol of AgNO3)

(2 mol of AgNo3) * (0.180884) / (1) = (? mol of AgNO3)

? = (0.361768 mol of AgNO3)

The question says what is the mass so we have to convert the answer we got into grams(g). In order to do that we need our formula:

mol = mass/molar mass

(0.361768 mol) = (g) / (169.87g/mol)

g = (0.361768 mol) * (169.87g/mol)

g = 61.45714 g

g = 61.5 g