Question

Heres a balanced equation:

Na2CO3(aq)+2AgNO3 (aq) --> 2NaNO3(aq)+Ag2CO3 (s)

From this equation, 10 mL of AgNO3 in this contains 34 grams of AgNO3. Calculate the grams of solid product, Ag2CO3.

Part 2:
10 mL of Na2CO3 in this experiment contains 21 grams of Na2CO3. Calculation that the grams of solid product, Ag2CO3, expected from this reaction.

(please show steps please, I need to understand how you got the answer, thanks)

Answer 1

mols Ag (part a) = g/molar mass = about 0.2 but you need to do it more accurately.
Using the coefficients in the balanced equation convert mols AgNO3 to mols Ag2CO3. That is 0.2 mol AgNO3 x (1 mol Ag2CO2/2 mols Ag_) = 0.2 x 1/2 = aboaut 0.1 mol.
g Ag2CO3 formed = mols x molar mass

b. Do the same and solve for mols (then grams) Ag2CO3.
If the question is asking for grams Ag2CO3 with 34g AgNO3 and 21g Na2CO3 it turns into a limiting reagent problem in which case the FEWER mols Ag2CO3 will be formed..