posted by Jabasha on .
if given 10.0grams of aluminum and 20.00 grams of silver carbonate how many grams would it take to make aluminum carbonate from silver carbonate
2Al + 3Ag2CO3 ==> Al2(CO3)3 + 6Ag
Convert 10 g Al to mols. mol = grams/molar mass. About 0.4
Convert 20 g Ag2CO3 to mols. About 0.07
Using the coefficients in the balanced equation, convert mol Al to mols
Al2(CO3)2. 0.4 x 1/2 = about 0.2
Same for Ag2CO3. About 0.07 x 1/3 = about 0.025
Both of these answers can't be right; the correct answer in limiting reagent problems is ALWAYS the smaller value and the reagent producing that value is the limiting reagent. The limiting reagent in this case is Ag2CO3 and you can produce approximately 0.025 mol Al2(CO3)2.
Is the question to determine how much Al is needed to react with the 20 g Ag2CO3? Then
0.07 mols Ag2CO3 x (2 mols Al/3 mols Ag2CO3) = 0.07 x 2/3 = about 0.05 mols Al needed. g Al = mols x molar mass = about 0.05 x 27 = about 1.35 g Al.
You need to redo all of this reading the calculator closer than I did.