Calculate the number of grams of calcium chloride (CaCl2) needed to react with excess silver nitrate (AgNO3) to produce 6.60 g of silver chloride (AgCl). The equation for the reaction is given below:
CaCl2 + 2AgNO3 → 2AgCl + Ca(NO3)2
To calculate the number of grams of calcium chloride (CaCl2) needed to react with excess silver nitrate (AgNO3) to produce 6.60 g of silver chloride (AgCl), we will use the stoichiometry of the balanced equation.
From the balanced equation: CaCl2 + 2AgNO3 → 2AgCl + Ca(NO3)2
We can see that 1 mole of CaCl2 reacts with 2 moles of AgNO3 to produce 2 moles of AgCl. Therefore, the molar ratio between CaCl2 and AgCl is 1:2.
Step 1: Calculate the molar mass of AgCl.
The molar mass of silver chloride (AgCl) can be calculated by adding the atomic masses of silver (Ag) and chlorine (Cl).
Ag: 1 x 107.87 g/mol = 107.87 g/mol
Cl: 1 x 35.45 g/mol = 35.45 g/mol
AgCl: 107.87 g/mol + 35.45 g/mol = 143.32 g/mol
Step 2: Calculate the number of moles of AgCl.
To find the number of moles, we divide the given mass of AgCl by its molar mass.
Moles of AgCl = mass of AgCl / molar mass of AgCl
Moles of AgCl = 6.60 g / 143.32 g/mol
Moles of AgCl = 0.0460 mol
Step 3: Use the stoichiometry to find the moles of CaCl2.
Since the molar ratio between CaCl2 and AgCl is 1:2, the number of moles of CaCl2 will be half of the moles of AgCl.
Moles of CaCl2 = 0.0460 mol / 2
Moles of CaCl2 = 0.0230 mol
Step 4: Calculate the number of grams of CaCl2.
To find the number of grams, we multiply the number of moles of CaCl2 by its molar mass.
Grams of CaCl2 = moles of CaCl2 x molar mass of CaCl2
Grams of CaCl2 = 0.0230 mol x 110.98 g/mol
Grams of CaCl2 = 2.55 g
Therefore, the number of grams of calcium chloride (CaCl2) needed to react with excess silver nitrate (AgNO3) to produce 6.60 g of silver chloride (AgCl) is 2.55 grams.
To find the number of grams of calcium chloride needed to react with excess silver nitrate, we need to use stoichiometry and the given equation for the reaction.
First, let's calculate the molar mass of silver chloride (AgCl). It consists of one silver atom (Ag) with a molar mass of 107.87 g/mol and one chlorine atom (Cl) with a molar mass of 35.45 g/mol. Adding these together, we get:
Molar mass of AgCl = 107.87 g/mol + 35.45 g/mol = 143.32 g/mol
Next, we can set up a stoichiometric ratio using the coefficients from the balanced equation. According to the balanced equation:
1 mole of CaCl2 reacts with 2 moles of AgNO3 to produce 2 moles of AgCl
Using this ratio, we can convert the given mass of silver chloride (AgCl) into moles:
Moles of AgCl = Mass of AgCl / Molar mass of AgCl
Moles of AgCl = 6.60 g / 143.32 g/mol ≈ 0.046 moles
Since the stoichiometry of the reaction tells us that 1 mole of CaCl2 reacts with 2 moles of AgNO3 to produce 2 moles of AgCl, we can say that:
1 mole of CaCl2 produces 2 moles of AgCl
Therefore, we can use the moles of AgCl to calculate the moles of CaCl2 needed:
Moles of CaCl2 = (Moles of AgCl / 2) ≈ 0.046 moles / 2 ≈ 0.023 moles
Now, to find the mass of calcium chloride (CaCl2) needed, we use the molar mass of CaCl2, which is 110.98 g/mol:
Mass of CaCl2 = Moles of CaCl2 × Molar mass of CaCl2
Mass of CaCl2 = 0.023 moles × 110.98 g/mol ≈ 2.56 g
Thus, approximately 2.56 grams of calcium chloride (CaCl2) are needed to react with excess silver nitrate (AgNO3) to produce 6.60 grams of silver chloride (AgCl).
6.6g AgCl = .046 moles
each mole CaCl2 yields 2 moles AgCl, so only 0.023 moles CaCl2 are needed.
.023 moles CaCl2 = 2.55g