Question

how many grams of silver chloride are produced when 45 g of calcium chloride react with excess silver nitrate?​

Answer 1

To solve this problem, we need to use the concept of stoichiometry to calculate the mass of silver chloride produced.

Step-by-step explanation:

To solve this problem, we need to use the concept of stoichiometry to calculate the mass of silver chloride produced. The balanced chemical equation for the reaction between calcium chloride (CaCl2) and silver nitrate (AgNO3) is:

CaCl2 + 2AgNO3 → 2AgCl + Ca(NO3)2

From this equation, we can see that the molar ratio between calcium chloride and silver chloride is 1:2. Therefore, we need to convert the mass of calcium chloride to moles using its molar mass, and then use the molar ratio to calculate the moles of silver chloride formed. Finally, we can convert the moles of silver chloride to grams using its molar mass.

First, calculate the moles of calcium chloride:

Moles of CaCl2 = mass of CaCl2 / molar mass of CaCl2

Molar mass of CaCl2 = atomic mass of calcium + 2 * atomic mass of chlorine = 40.08 g/mol + 2 * 35.45 g/mol

Next, calculate the moles of silver chloride:

Moles of AgCl = moles of CaCl2 * (2 moles of AgCl / 1 mole of CaCl2)

Finally, convert the moles of silver chloride to grams:

Mass of AgCl = moles of AgCl * molar mass of AgCl = 2 * (107.87 g/mol)

Now plug in the given values:

Mass of CaCl2 = 45 g

Molar mass of CaCl2 = 40.08 g/mol + 2 * 35.45 g/mol

Molar mass of AgCl = 2 * 107.87 g/mol

Moles of CaCl2 = 45 g / (40.08 g/mol + 2 * 35.45 g/mol)

Moles of AgCl = moles of CaCl2 * (2 moles of AgCl / 1 mole of CaCl2)

Mass of AgCl = moles of AgCl * molar mass of AgCl

Solve the equations to find the mass of AgCl:

Mass of AgCl = (45 g / (40.08 g/mol + 2 * 35.45 g/mol)) * (2 * 107.87 g/mol) = 85.85 g

Therefore, when 45 g of calcium chloride reacts with excess silver nitrate, approximately 85.85 g of silver chloride are produced.

Answer 2

= 116 g AgCl

Step-by-step explanation:

CaCl2 + 2AgNO3 = Ca(NO3)2 + 2AgCl

Mass of AgCl;

1 Mole of CaCl2 = 111 g

Moles of CaCl2;

= 45 g CaCl2 (1 mol CaCl2/111 g CaCl2)

= 0.405 moles

Mole Ratio = 2: 1

Therefore; moles ,of AgCl

= 0.405 moles × 2

= 0.810 moles

Thus;

Mass of AgCl = 0.810 moles × 143.5 g AgCl

= 116 g AgCl