If you mix 40.0 mL of a 0.389 M solution of K2CrO4 with 40.0 mL of a 0.389 M solution of AgNO3, what mass of solid forms?
To determine the mass of solid that forms when mixing these solutions, we need to identify the limiting reagent first. The limiting reagent is the reactant that is completely consumed and determines the maximum amount of product that can be formed.
1. Calculate the moles of each reactant:
Moles of K2CrO4 = volume (in L) x molarity
Moles of K2CrO4 = 0.040 L x 0.389 mol/L = 0.01556 mol
Moles of AgNO3 = volume (in L) x molarity
Moles of AgNO3 = 0.040 L x 0.389 mol/L = 0.01556 mol
2. Use the balanced chemical equation to determine the stoichiometry between the reactants and the product.
The balanced equation for the reaction between K2CrO4 and AgNO3 is:
2 K2CrO4 (aq) + 3 AgNO3 (aq) -> Ag2CrO4 (s) + 2 KNO3 (aq)
According to the balanced equation, 2 moles of K2CrO4 react with 3 moles of AgNO3 to form 1 mole of Ag2CrO4.
3. Determine the limiting reagent:
Since the moles of K2CrO4 and AgNO3 are equal (0.01556 mol), neither is present in excess. However, we see that 2 moles of K2CrO4 react with 3 moles of AgNO3, so the reactant that has a smaller stoichiometric ratio is the limiting reagent. In this case, K2CrO4 is the limiting reagent.
4. Calculate the moles of Ag2CrO4 formed:
From the balanced equation, we know that 2 moles of K2CrO4 react to form 1 mole of Ag2CrO4.
Therefore, moles of Ag2CrO4 = (moles of K2CrO4) x (1 mole Ag2CrO4 / 2 moles K2CrO4)
moles of Ag2CrO4 = 0.01556 mol x (1 mol / 2 mol) = 0.00778 mol
5. Determine the molar mass of Ag2CrO4:
Ag2CrO4 contains two silver (Ag) atoms with a molar mass of approximately 107.8682 g/mol and one chromate (CrO4) ion with a molar mass of approximately 115.9952 g/mol. Calculating the molar mass:
Molar mass of Ag2CrO4 = (2 x molar mass of Ag) + molar mass of CrO4
= (2 x 107.8682 g/mol) + 115.9952 g/mol
= 231.7316 g/mol
6. Calculate the mass of Ag2CrO4 formed:
Mass of Ag2CrO4 = moles of Ag2CrO4 x molar mass of Ag2CrO4
= 0.00778 mol x 231.7316 g/mol
= 1.804 g (rounded to 3 decimal places)
Therefore, the mass of solid Ag2CrO4 formed when mixing 40.0 mL of a 0.389 M solution of K2CrO4 with 40.0 mL of a 0.389 M solution of AgNO3 is approximately 1.804 grams.
To determine the mass of solid formed when mixing the two solutions, we need to find the limiting reagent. The limiting reagent is the reactant that is completely consumed and determines the amount of product formed.
First, we need to write the balanced equation for the reaction between K2CrO4 and AgNO3:
2 K2CrO4 + 2 AgNO3 → Ag2CrO4 + 2 KNO3
From the balanced equation, we can see that two moles of K2CrO4 react with two moles of AgNO3 to form one mole of Ag2CrO4.
Next, let's determine the number of moles of K2CrO4 and AgNO3 we have:
Moles of K2CrO4 = (volume in liters) x (molarity)
= (40.0 mL / 1000 mL/L) x (0.389 mol/L)
= 0.01556 mol
Moles of AgNO3 = (volume in liters) x (molarity)
= (40.0 mL / 1000 mL/L) x (0.389 mol/L)
= 0.01556 mol
Since the stoichiometry of the reaction indicates that the reactants are in a 1:1 ratio, this means that both reactants are present in equimolar amounts.
Now, let's determine the molar mass of Ag2CrO4:
Molar mass of Ag2CrO4 = (2 x atomic mass of Ag) + atomic mass of Cr + (4 x atomic mass of O)
= (2 x 107.87 g/mol) + 52.01 g/mol + (4 x 16.00 g/mol)
= 331.75 g/mol
To find the mass of Ag2CrO4 formed, we can use the equation:
Mass of Ag2CrO4 = Moles of Ag2CrO4 x Molar mass of Ag2CrO4
Mass of Ag2CrO4 = 0.01556 mol x 331.75 g/mol
= 5.15 g
Therefore, when mixing 40.0 mL of a 0.389 M solution of K2CrO4 with 40.0 mL of a 0.389 M solution of AgNO3, approximately 5.15 grams of solid Ag2CrO4 will form.
Here is an example of a stoichiometry problem. Just follow the steps.
http://www.jiskha.com/science/chemistry/stoichiometry.html