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.
To determine the amount of grams required to make aluminum carbonate from silver carbonate, we need to consider the balanced chemical equation and use stoichiometry.
The balanced chemical equation for the reaction between silver carbonate (Ag2CO3) and aluminum (Al) is:
2Ag2CO3 + 3Al → Al2(CO3)3 + 4Ag
From this equation, we can see that it takes 2 moles of silver carbonate to produce 1 mole of aluminum carbonate. Therefore, we need to convert the given masses of aluminum and silver carbonate into moles to determine the mole ratio.
Step 1: Convert the mass of aluminum into moles.
Given mass of aluminum = 10.0 grams
The molar mass of aluminum (Al) is approximately 26.98 g/mol.
Number of moles of aluminum = given mass / molar mass
Number of moles of aluminum = 10.0 g / 26.98 g/mol
Step 2: Convert the mass of silver carbonate into moles.
Given mass of silver carbonate = 20.0 grams
The molar mass of silver carbonate (Ag2CO3) is approximately 275.74 g/mol.
Number of moles of silver carbonate = given mass / molar mass
Number of moles of silver carbonate = 20.0 g / 275.74 g/mol
Step 3: Determine the mole ratio between silver carbonate and aluminum carbonate.
From the balanced chemical equation, we know that 2 moles of Ag2CO3 produce 1 mole of Al2(CO3)3.
Step 4: Calculate the moles of aluminum carbonate produced.
Number of moles of aluminum carbonate = Number of moles of silver carbonate / 2
Step 5: Convert the moles of aluminum carbonate into grams.
Given that the molar mass of aluminum carbonate (Al2(CO3)3) is approximately 233.99 g/mol, we can use the equation:
Mass of aluminum carbonate = number of moles * molar mass
Mass of aluminum carbonate = Number of moles of aluminum carbonate * 233.99 g/mol
This gives us the amount of grams of aluminum carbonate produced from the reaction between the given masses of aluminum and silver carbonate.