2.7g of Ag2Co3 on being heated strongly yields a residue weighing.......
To determine the weight of the residue when 2.7g of Ag2CO3 is heated strongly, we need to understand the chemical reaction that occurs.
Ag2CO3 can be decomposed into Ag2O (silver(I) oxide) and CO2 (carbon dioxide) when heated:
2 Ag2CO3(s) → 4 Ag2O(s) + 2 CO2(g)
The molar mass of Ag2CO3 is calculated as follows:
Ag = 107.87 g/mol
C = 12.01 g/mol
O = 16.00 g/mol
Total molar mass of Ag2CO3 = 2(2 × Ag) + C + 3(2 × O) = 2(2 × 107.87) + 12.01 + 3(2 × 16.00) = 275.71 g/mol
Now, we can calculate the number of moles of Ag2CO3 in 2.7g using the formula:
Number of moles = Mass / Molar mass
Number of moles of Ag2CO3 = 2.7g / 275.71 g/mol ≈ 0.0098 mol
From the balanced chemical equation, we can see that 2 moles of Ag2CO3 yield 4 moles of Ag2O.
So, 0.0098 mol of Ag2CO3 will yield (4/2) × 0.0098 mol = 0.0196 mol of Ag2O.
The molar mass of Ag2O is calculated as follows:
2 × Ag = 2 × 107.87 g/mol = 215.74 g/mol
O = 16.00 g/mol
Total molar mass of Ag2O = 2 × Ag + O = 215.74 + 16.00 = 231.74 g/mol
To determine the weight of the residue (Ag2O), we use the formula:
Weight = Number of moles × Molar mass
Weight of Ag2O = 0.0196 mol × 231.74 g/mol ≈ 4.54 g
Therefore, when 2.7g of Ag2CO3 is heated strongly, it yields a residue (Ag2O) weighing approximately 4.54g.
2.11 g
2.27
Ag2CO3 ==> Ag2O + CO2
Convert 2.7 g Ag2CO3 to mols. moles = grams/molar mass.
Using the coefficient in the balanced equation, convert moles Ag2CO3 to moles Ag2O.
Now convert moles Ag2O to grams. g = moles x molar mass.