An excess of aqueous AgNO3 reacts with 43 mL of 5 M K2CrO4(aq)to form a precipitate. what is it? What mass of precipitate is formed?
answer in g
2AgNO3 + K2Cr2O7 ==> Ag2Cr2O7 + 2KNO3
mols K2Cr2O7 = M x L = ?
mols Ag2Cr2O7 =mols K2Cr2O7
Then g Ag2Cr2O7 = mols Ag2Cr2O7 x molar mass A2Cr2O7
Oh, chemistry! I bet AgNO3 and K2CrO4 are having quite the party. So, when you mix those two together, they have a little dance and form a lovely precipitate known as silver chromate (Ag2CrO4). Isn't that fancy?
Now, about the mass of the precipitate... To figure that out, we need to do some calculations.
First, let's find out the number of moles of K2CrO4:
43 mL of 5 M K2CrO4 gives us:
43 mL * (5 mol/L) = 215 mol
Since AgNO3 and K2CrO4 have a stoichiometric ratio of 2:1, we can say that we'll form half the number of moles of Ag2CrO4.
That means, the moles of Ag2CrO4 formed will be:
0.5 * 215 mol = 107.5 mol
The molar mass of Ag2CrO4 is approximately 332.1 g/mol. So, to find the mass of the precipitate, we can multiply the number of moles by the molar mass:
Mass of precipitate = 107.5 mol * 332.1 g/mol
= 35,692.5 g
That's quite a lot of precipitate! A silver chromate party indeed!
To determine the precipitate formed and its mass, we need to consider the chemical reaction between AgNO3 and K2CrO4:
2AgNO3(aq) + K2CrO4(aq) -> Ag2CrO4(s) + 2KNO3(aq)
From the balanced equation, we can see that Ag2CrO4 is the precipitate formed.
To find the mass of the precipitate, we need to calculate the moles of AgNO3 and K2CrO4 used, and then determine the limiting reagent.
Step 1: Calculate the moles of K2CrO4 used:
Molarity of K2CrO4 = 5 M
Volume of K2CrO4 = 43 mL = 0.043 L
Moles of K2CrO4 = Molarity * Volume
= 5 M * 0.043 L
= 0.215 mol
Step 2: Calculate the moles of AgNO3 used:
From the balanced equation, we can see that the ratio of AgNO3 to K2CrO4 is 2:1.
Hence, the moles of AgNO3 used will be half the moles of K2CrO4 used.
Moles of AgNO3 = (0.215 mol)/2
= 0.1075 mol
Step 3: Determine the limiting reagent:
The limiting reagent is the one that produces the least amount of product. To find this, we compare the moles of AgNO3 and K2CrO4 used.
Moles of AgNO3 = 0.1075 mol
Moles of K2CrO4 = 0.215 mol
Since the ratio of AgNO3 to K2CrO4 is 1:2, it means that for every 0.1075 moles of AgNO3, we need 0.215 moles of K2CrO4. In this case, we have enough K2CrO4 to react fully with AgNO3, and AgNO3 is the limiting reagent.
Step 4: Calculate the mass of Ag2CrO4 precipitate formed:
From the balanced equation, we know that the ratio of AgNO3 to Ag2CrO4 is 2:1.
So, the moles of Ag2CrO4 formed will be half the moles of AgNO3 used.
Moles of Ag2CrO4 = (0.1075 mol)/2
= 0.05375 mol
To calculate the mass of Ag2CrO4, we need to multiply the moles by its molar mass.
Molar mass of Ag2CrO4 = (2 * atomic mass of Ag) + atomic mass of Cr + (4 * atomic mass of O)
= (2 * 107.87 g/mol) + 52.00 g/mol + (4 * 16.00 g/mol)
= 316.74 g/mol
Mass of Ag2CrO4 = Moles * Molar mass
= 0.05375 mol * 316.74 g/mol
= 17.0196 g
Therefore, the mass of Ag2CrO4 precipitate formed is 17.0196 grams.
To determine the precipitate formed and its mass, we need to consider the chemical reaction between the reactants, silver nitrate (AgNO3) and potassium chromate (K2CrO4). The balanced chemical equation for this reaction is:
2 AgNO3 + K2CrO4 → Ag2CrO4 + 2 KNO3
From the balanced equation, we can deduce that silver chromate (Ag2CrO4) is the precipitate formed.
Now, to calculate the mass of precipitate formed, we need to use the given information about the volume and concentration of the potassium chromate solution.
Given:
Volume of K2CrO4 solution (V) = 43 mL = 0.043 L
Concentration of K2CrO4 solution (C) = 5 M
To find the number of moles of K2CrO4 used, we can use the formula:
moles = volume (L) * concentration (M)
moles of K2CrO4 = V * C
= 0.043 L * 5 M
= 0.215 moles
From the balanced chemical equation, we see that the ratio between the moles of K2CrO4 used and Ag2CrO4 produced is 1:1. Thus, the number of moles of Ag2CrO4 formed is also 0.215 moles.
Next, we need to convert moles of Ag2CrO4 into grams by using its molar mass (formula weight). The molar mass of Ag2CrO4 is:
Ag2CrO4 = (2 x atomic mass of Ag) + atomic mass of Cr + (4 x atomic mass of O)
= (2 x 107.87 g/mol) + 52.00 g/mol + (4 x 16.00 g/mol)
= 331.74 g/mol
Now, we can calculate the mass of Ag2CrO4:
mass = moles * molar mass
= 0.215 moles * 331.74 g/mol
≈ 71.38 g
Therefore, approximately 71.38 grams of Ag2CrO4 precipitate is formed.