Silver chromate may be prepared by the following reaction.
2 AgNO3(aq) + K2CrO4(aq) Ag2CrO4(s) + 2 KNO3(aq)
Calculate the mass of silver chromate produced when 0.623 L of 1.345 M silver nitrate is reacted with 245 mL of 1.89 M potassium chromate.
This is a limiting regent (LR) problem. You know that when amounts are given for BOTH reactants.
mols AgNO3 = M x L = ?
mols K2C4O4 = M x L = ?
Using the coefficients in the balanced equation convert mols AgNO3 to mols Ag2CrO4
Do the same for mols K2CrO4 to mols Ag2CrO4
It is likely that these two values for mols Ag2CrO4 don't agree which means one is not right. The correct value in LR problems is ALWAYS the smaller value and the reagent producing that value is the LR.
Now take the smaller value of mols and convert to grams. g = mols x molar mass
To calculate the mass of silver chromate produced, we need to use the stoichiometry of the reaction and the given concentrations and volumes.
First, let's convert the volume of the silver nitrate solution to moles:
0.623 L × 1.345 mol/L = 0.837 mol AgNO3
Next, convert the volume of the potassium chromate solution to moles:
245 mL × 0.001 L/mL × 1.89 mol/L = 0.464 mol K2CrO4
According to the balanced equation, the mole ratio between AgNO3 and Ag2CrO4 is 2:1. Therefore, the moles of Ag2CrO4 produced will also be 0.837 mol.
Using the molar mass of silver chromate (Ag2CrO4), which can be found by adding the atomic masses of its constituent elements:
Ag: 2 × 107.87 g/mol = 215.74 g/mol
Cr: 1 × 52.00 g/mol = 52.00 g/mol
O: 4 × 16.00 g/mol = 64.00 g/mol
Total molar mass = 215.74 g/mol + 52.00 g/mol + 64.00 g/mol = 331.74 g/mol
Finally, calculate the mass of silver chromate produced:
Mass = Moles × Molar mass
= 0.837 mol × 331.74 g/mol
= 277.61 g
Therefore, the mass of silver chromate produced is 277.61 grams.
To calculate the mass of silver chromate produced, we need to follow these steps:
Step 1: Convert the volume of the solutions to the corresponding number of moles using the given concentrations.
Given:
Volume of silver nitrate solution (V1) = 0.623 L
Concentration of silver nitrate solution (C1) = 1.345 M
Number of moles of silver nitrate (n1) = Volume x Concentration
n1 = V1 x C1
Step 2: Follow the same steps for the potassium chromate solution.
Given:
Volume of potassium chromate solution (V2) = 245 mL = 0.245 L
Concentration of potassium chromate solution (C2) = 1.89 M
Number of moles of potassium chromate (n2) = Volume x Concentration
n2 = V2 x C2
Step 3: Determine the limiting reactant.
To determine the limiting reactant, we compare the number of moles of each reactant. The reactant that produces the smaller amount of the product will be the limiting reactant.
Step 4: Calculate the number of moles of silver chromate formed.
The balanced equation shows that the stoichiometric ratio between silver nitrate and silver chromate is 2:1. This means that for every 2 moles of silver nitrate, 1 mole of silver chromate will be produced.
Number of moles of silver chromate (n3) = (n1 / 2) or (n2 / 1)
Step 5: Convert moles of silver chromate to mass.
To calculate the mass of silver chromate, we use its molar mass. The molar mass of silver chromate (Ag2CrO4) can be calculated by adding the atomic masses of its constituent elements.
Molar mass of Ag2CrO4 = (2 x atomic mass of Ag) + atomic mass of Cr + (4 x atomic mass of O)
Step 6: Calculate the mass of silver chromate.
Mass of silver chromate (m) = Number of moles (n3) x Molar mass
Following these steps should help you calculate the mass of silver chromate produced.