7.82 SILVER(I) NITRATE REACT WITH POTASSIUM CHLORIDE FIND THE BALANCED CHEMICAL REACTION

USING THE GIVEN MASS FIND THE MASS OF THE REAGENT. ALSO CALCULATE THE MASSES OF THE PRODOUCT FORMED IN THE REACTION

To balance the chemical reaction between silver nitrate (AgNO3) and potassium chloride (KCl), we need to ensure that the number of atoms of each element on both sides of the reaction is the same. Here's how you can balance the equation:

AgNO3 + KCl -> AgCl + KNO3

Now, let's find the mass of the reagent (silver nitrate) using the given mass.

Step 1: Determine the molar mass of silver nitrate (AgNO3). The molar mass can be calculated by adding the atomic masses of silver (Ag), nitrogen (N), and three oxygen (O) atoms.

Ag: 107.87 g/mol
N: 14.01 g/mol
O: 16.00 g/mol (x3) = 48.00 g/mol

So, the molar mass of AgNO3 is:
107.87 g/mol + 14.01 g/mol + 48.00 g/mol = 169.88 g/mol

Step 2: Use the molar mass to find the mass of silver nitrate (AgNO3) using the given mass.

Let's assume the given mass of silver nitrate is 7.82 grams.

Number of moles = Given mass / molar mass
Number of moles of AgNO3 = 7.82 g / 169.88 g/mol ≈ 0.046 moles

Step 3: Now, let's calculate the masses of the products formed in the reaction.

From the balanced equation, we see that 1 mole of AgNO3 produces 1 mole of AgCl and 1 mole of KNO3.

The molar mass of AgCl is:
Ag: 107.87 g/mol + Cl: 35.45 g/mol = 143.32 g/mol

The molar mass of KNO3 is:
K: 39.10 g/mol + N: 14.01 g/mol + O: 16.00 g/mol (x3) = 101.10 g/mol

Now we can calculate the masses using the number of moles obtained earlier:

Mass of AgCl = 0.046 moles × 143.32 g/mol ≈ 6.60 grams
Mass of KNO3 = 0.046 moles × 101.10 g/mol ≈ 4.65 grams

Therefore, the balanced chemical reaction is:
AgNO3 + KCl -> AgCl + KNO3

The mass of the reagent (silver nitrate) is approximately 7.82 grams. The masses of the products formed in the reaction are approximately 6.60 grams of AgCl and 4.65 grams of KNO3.