A mixture is prepared by adding 100 ml of 0.10M Na2CrO4 to 100.ml of ).10 N AgNO3. A precipitate forms in this mixture. The precipitate is separated from the mixture by filtration.
Question: What quantity of solid product is produced?
To determine the quantity of solid product produced in this reaction, we need to find out the limiting reagent, which is the reactant that is completely consumed and determines the maximum amount of product that can be formed.
Let's start by writing down the balanced chemical equation for the reaction:
2 AgNO3 + Na2CrO4 -> Ag2CrO4 + 2 NaNO3
From the equation, we can see that 2 moles of AgNO3 react with 1 mole of Na2CrO4 to produce 1 mole of Ag2CrO4.
Next, let's calculate the number of moles for each reactant:
Number of moles of Na2CrO4: 0.10 M * 0.100 L = 0.010 moles
Number of moles of AgNO3: 0.10 M * 0.100 L = 0.010 moles
From the calculation, we can see that both reactants have the same number of moles. Therefore, neither is in excess, and they are both the limiting reagents.
Since both reactants are used in a 1:1 ratio to form the product, the number of moles of Ag2CrO4 produced will be equal to the number of moles of the limiting reagent.
So, 0.010 moles of Ag2CrO4 will be produced.
To find the quantity of solid product produced, we need to multiply the number of moles by its molar mass. The molar mass of Ag2CrO4 can be calculated as follows:
Molar mass of Ag = 107.87 g/mol (atomic mass of silver)
Molar mass of Cr = 52.00 g/mol (atomic mass of chromium)
Molar mass of O = 16.00 g/mol (atomic mass of oxygen)
Molar mass of Ag2CrO4 = 2 * molar mass of Ag + molar mass of Cr + 4 * molar mass of O
= 2 * 107.87 + 52.00 + 4 * 16.00
= 331.74 g/mol
Now, we can calculate the quantity of solid product:
Quantity of solid product = Number of moles * Molar mass
= 0.010 moles * 331.74 g/mol
= 3.32 grams
Therefore, the quantity of solid product produced is 3.32 grams.