2AgNO3 + CaCl2 ---> Ca(NO3)2 + 2AgCl
all of the substances involved in this reaction are soluble in water except AgCl which forms a solid at the bottom of the flask . Suppose we mix together a solution containing 12.6 g of AgNO3 and one containing 8.4 g of CaCl2 .What mass of AgCl is formed?
This is a limiting reagent (LR) problem.
mols AgNO3 = grams/molar mass = ?
mols CaCl2 = grams/molar mass = ?
Using the coefficients in the balanced equation, convert mols AgNO3 to mols AgCl.
Do the same to convert mols CaCl2 to mols AgCl.
It is likely the two values for mols AgCl will not agree which means both answers can't be right. The correct value in LR problems is ALWAYS the smaller value. Choose that number and convert to grams AgCl. grams AgCl = mols AgCl x molar mass AgCl.
To determine the mass of AgCl formed in the reaction, we need to follow these steps:
Step 1: Calculate the molar mass of AgNO3:
AgNO3: Ag (1 × 107.87 g/mol) + N (1 × 14.01 g/mol) + O (3 × 16.00 g/mol)
AgNO3: 107.87 g/mol + 14.01 g/mol + 48.00 g/mol = 169.88 g/mol
Step 2: Calculate the moles of AgNO3:
moles of AgNO3 = mass / molar mass
moles of AgNO3 = 12.6 g / 169.88 g/mol ≈ 0.0743 mol
Step 3: Calculate the molar mass of CaCl2:
CaCl2: Ca (1 × 40.08 g/mol) + Cl (2 × 35.45 g/mol)
CaCl2: 40.08 g/mol + 70.90 g/mol = 110.98 g/mol
Step 4: Calculate the moles of CaCl2:
moles of CaCl2 = mass / molar mass
moles of CaCl2 = 8.4 g / 110.98 g/mol ≈ 0.0756 mol
Step 5: Determine the limiting reactant:
The reactant that produces the lesser amount of product is the limiting reactant. In this case, AgNO3 and CaCl2 have a 1:1 stoichiometric ratio, so the limiting reactant will be the one with fewer moles. In this case, AgNO3 has fewer moles (0.0743 mol), so it is the limiting reactant.
Step 6: Calculate the moles of AgCl formed:
From the balanced equation: 2 moles of AgNO3 produce 2 moles of AgCl
moles of AgCl formed = moles of AgNO3 × (2 moles of AgCl / 2 moles of AgNO3) = 0.0743 mol
Step 7: Calculate the mass of AgCl formed:
mass of AgCl formed = moles of AgCl formed × molar mass of AgCl
mass of AgCl formed = 0.0743 mol × (107.87 g/mol) ≈ 8.01 g
Therefore, approximately 8.01 grams of AgCl are formed in the reaction.
To find the mass of AgCl formed, we need to determine the limiting reactant in the reaction. The limiting reactant is the one that is completely consumed, thereby determining the maximum amount of product that can be formed.
First, let's calculate the number of moles for each reactant:
For AgNO3:
Molar mass of AgNO3 = 107.87 g/mol + 14.01 g/mol + (3 * 16.00 g/mol) = 169.87 g/mol
Number of moles of AgNO3 = mass / molar mass = 12.6 g / 169.87 g/mol
For CaCl2:
Molar mass of CaCl2 = 40.08 g/mol + (2 * 35.45 g/mol) = 110.98 g/mol
Number of moles of CaCl2 = mass / molar mass = 8.4 g / 110.98 g/mol
Next, we need to determine the stoichiometric ratio between AgNO3 and AgCl. From the balanced chemical equation, we can see that 2 moles of AgNO3 react with 2 moles of AgCl.
By comparing the moles of AgNO3 and CaCl2, we can see that the stoichiometric ratio is 2:1. This means that 2 moles of AgNO3 will form 1 mole of AgCl.
To find the limiting reactant, we compare the moles of AgNO3 and AgCl formed. Since 2 moles of AgNO3 are required to form 1 mole of AgCl, we divide the number of moles of AgNO3 by 2:
Moles of AgCl formed = (moles of AgNO3) / 2
Now, let's plug in the values:
Moles of AgNO3 = 12.6 g / 169.87 g/mol
Moles of AgCl formed = (12.6 g / 169.87 g/mol) / 2
Now, we can calculate the mass of AgCl formed using the molar mass of AgCl:
Molar mass of AgCl = 107.87 g/mol + 35.45 g/mol = 143.32 g/mol
Mass of AgCl formed = (moles of AgCl formed) * (molar mass of AgCl)
Finally, let's substitute the values and calculate:
Mass of AgCl formed = ((12.6 g / 169.87 g/mol) / 2) * 143.32 g/mol
Now you can solve the equation to find the mass of AgCl formed.