When silver nitrate reacts with barium chloride, silver chloride and barium nitrate are formed. How many grams of silver chloride are formed when 8.0 g of silver nitrate reacts with 15.0 g of barium chloride?

Please explain how to do this. I honestly do not know.

To determine the amount of silver chloride formed when 8.0 g of silver nitrate (AgNO3) reacts with 15.0 g of barium chloride (BaCl2), we need to follow these steps:

1. Write the balanced chemical equation for the reaction:
AgNO3 + BaCl2 -> AgCl + Ba(NO3)2

2. Calculate the moles of each substance:
Moles of AgNO3 = mass / molar mass of AgNO3
= 8.0 g / (107.87 g/mol + 14.01 g/mol + 3(16.00 g/mol))
≈ 0.0496 mol

Moles of BaCl2 = mass / molar mass of BaCl2
= 15.0 g / (137.33 g/mol + 2(35.45 g/mol))
≈ 0.0797 mol

3. Determine the limiting reactant:
The limiting reactant is the one that is completely consumed in the reaction, thereby determining the maximum amount of product formed. To do this, we compare the mole ratio of AgNO3 to BaCl2 in the balanced equation.

From the balanced equation:
2 moles of AgNO3 react with 1 mole of BaCl2

Moles of BaCl2 required = 2 * moles of AgNO3
≈ 2 * 0.0496 mol
≈ 0.0993 mol

Since we have only 0.0797 mol of BaCl2, it is the limiting reactant.

4. Calculate the moles of AgCl formed:
Since the limiting reactant is BaCl2, we use its moles to calculate the moles of AgCl formed. According to the balanced equation, 1 mole of BaCl2 reacts to form 1 mole of AgCl.

Moles of AgCl formed ≈ moles of BaCl2
≈ 0.0797 mol

5. Calculate the mass of AgCl formed:
Mass of AgCl = moles of AgCl formed * molar mass of AgCl
= 0.0797 mol * (107.87 g/mol + 35.45 g/mol)
≈ 12.75 g

Therefore, approximately 12.75 grams of silver chloride are formed when 8.0 grams of silver nitrate reacts with 15.0 grams of barium chloride.

To solve this problem, we need to use the concept of stoichiometry and the given mass of the reactants. Here's a step-by-step explanation of how to solve it:

Step 1: Write the balanced chemical equation for the reaction:

2 AgNO3 + BaCl2 → 2 AgCl + Ba(NO3)2

Step 2: Determine the molar masses of the substances involved:

AgNO3: 107.87 g/mol
BaCl2: 208.23 g/mol
AgCl: 143.32 g/mol

Step 3: Calculate the number of moles of silver nitrate and barium chloride using their molar masses:

moles of AgNO3 = 8.0 g / 107.87 g/mol = 0.0741 mol
moles of BaCl2 = 15.0 g / 208.23 g/mol = 0.0720 mol

Step 4: Determine the limiting reactant. This can be done by comparing the mole ratios of the reactants in the balanced equation. The reactant that has the smaller mole ratio is the limiting reactant.

From the balanced equation, the mole ratio of AgNO3 to AgCl is 2:2, while the mole ratio of BaCl2 to AgCl is 1:2. This means that for every 2 moles of AgNO3, we need 1 mole of BaCl2 to react completely.

Compared to the stoichiometric ratios, we have an excess of BaCl2 (0.0720 mol), as we only need 0.0741/2 = 0.037 mol of BaCl2 to react with 0.0741 mol of AgNO3. Therefore, AgNO3 is the limiting reactant.

Step 5: Calculate the moles of silver chloride that can be formed using the mole ratio from the balanced equation:

moles of AgCl = 0.0741 mol AgNO3 × (2 mol AgCl / 2 mol AgNO3) = 0.0741 mol

Step 6: Determine the mass of silver chloride using its molar mass:

mass of AgCl = 0.0741 mol × 143.32 g/mol = 10.63 g

Answer: When 8.0 g of silver nitrate reacts with 15.0 g of barium chloride, 10.63 g of silver chloride are formed.

Look in your text/notes about how to solve limiting reagent problems.