Solutions of sodium carbonate and silver nitrate react to form solid silver carbonate and a solution of sodium nitrate. A solution containing 0.55 mole of sodium carbonate is mixed with one containing 0.55 mole of silver nitrate. After the reaction is complete, the solutions are evaporated to dryness, leaving a mixture of salts. How many grams of silver carbonate are present after the reaction is complete?
Na2CO3 + 2AgNO3 = 2NaNO3 + Ag2CO3
each mole of Na2CO3 requires 2 moles of AgNO3, so only half the Na2CO3 is used up.
So, only 0.275 moles of Ag2CO3 are produced, or .275*275.7453 = 75.8g
To find the number of grams of silver carbonate formed, we need to use stoichiometry and the molar ratios between the reactants and products.
First, let's write the balanced equation for the reaction:
2 AgNO3 + Na2CO3 → Ag2CO3 + 2 NaNO3
From the balanced equation, we can see that 2 moles of silver nitrate react with 1 mole of sodium carbonate to produce 1 mole of silver carbonate.
Given that both the sodium carbonate and silver nitrate solutions contain 0.55 moles each, we can see that there is an equal number of moles for both reactants. Therefore, the limiting reactant is sodium carbonate.
Since the balanced equation shows a 1:1 mole ratio between sodium carbonate and silver carbonate, we can conclude that 0.55 moles of silver carbonate are formed.
Now, to convert moles to grams, we need to use the molar mass of silver carbonate (Ag2CO3), which is the sum of the atomic masses of its components. The molar mass of silver (Ag) is 107.87 g/mol, and the molar mass of carbonate (CO3) is 60.01 g/mol.
The molar mass of Ag2CO3 = 2(Ag) + 1(C) + 3(O)
= 2(107.87 g/mol) + 1(12.01 g/mol) + 3(16.00 g/mol)
= 275.73 g/mol
Now we can calculate the mass of silver carbonate:
Mass = moles × molar mass
= 0.55 mol × 275.73 g/mol
= 151.65 g
Therefore, after the reaction is complete, there are approximately 151.65 grams of silver carbonate present.