2AgNO3 + BaCl2 ==> 2AgCl + Ba(NO3)2

How many grams of silver chloride are produced from 5 g of silver nitrate reacting with an excess of barium chloride?

When silver nitrate reacts with sodium chloride, to form

silver chloride, the equivalent weight of AgNO3

To determine the amount of silver chloride produced, we need to first calculate the number of moles of silver nitrate in 5 g. Then, using the stoichiometry of the balanced equation, we can determine the number of moles of silver chloride produced. Finally, we use the molar mass of silver chloride to convert moles to grams.

Step 1: Calculate the moles of AgNO3
Molar mass of AgNO3 = atomic mass of Ag + atomic mass of N + 3 * atomic mass of O
= 108 + 14 + (3 * 16)
= 108 + 14 + 48
= 170 g/mol

Moles of AgNO3 = mass of AgNO3 / molar mass of AgNO3
= 5 g / 170 g/mol
≈ 0.0294 mol

Step 2: Calculate the moles of AgCl produced using stoichiometry
From the balanced equation, we see that 2 moles of AgNO3 produce 2 moles of AgCl.
So, 0.0294 mol of AgNO3 will produce 0.0294 mol of AgCl.

Step 3: Calculate the grams of AgCl produced
Molar mass of AgCl = atomic mass of Ag + atomic mass of Cl
= 108 + 35.5
= 143.5 g/mol

Grams of AgCl = moles of AgCl * molar mass of AgCl
= 0.0294 mol * 143.5 g/mol
≈ 4.23 g

Therefore, approximately 4.23 grams of silver chloride are produced from 5 grams of silver nitrate reacting with an excess of barium chloride.

To determine the number of grams of silver chloride produced, we need to calculate the theoretical yield of silver chloride. The theoretical yield represents the maximum amount of product that can be obtained from a given amount of reactant.

To find the theoretical yield, we need to use the balanced chemical equation and the molar masses of the reactants and products.

First, let's calculate the molar mass of silver nitrate (AgNO3):
AgNO3: Ag (atomic mass = 107.87 g/mol) + N (atomic mass = 14.01 g/mol) + 3O (atomic mass = 16.00 g/mol) = 169.87 g/mol

According to the balanced equation, 2 moles of silver nitrate (AgNO3) reacts with 1 mole of silver chloride (AgCl). This means that the molar ratio between AgNO3 and AgCl is 2:2.

Next, let's calculate the moles of silver nitrate:
moles of AgNO3 = mass of AgNO3 / molar mass of AgNO3
moles of AgNO3 = 5 g / 169.87 g/mol = 0.0294 mol

Since the molar ratio between AgNO3 and AgCl is 2:2, the moles of AgCl produced is also 0.0294 mol.

Finally, we calculate the grams of silver chloride:
mass of AgCl = moles of AgCl * molar mass of AgCl
mass of AgCl = 0.0294 mol * (Ag (atomic mass = 107.87 g/mol) + Cl (atomic mass = 35.45 g/mol)) = 3.41 g

Therefore, 3.41 grams of silver chloride are produced when 5 grams of silver nitrate reacts with an excess of barium chloride.

Here is a step by step procedure for these problems.

http://www.jiskha.com/science/chemistry/stoichiometry.html