How many grams of silver chloride can be produced if you start with 4.62 grams of barium chloride ? 2AgNO3 + BaCl2 —> 2AgCl + Ba(NO3)

See your post above.

6.36 grams

To determine the number of grams of silver chloride produced, we need to use stoichiometry. Stoichiometry allows us to calculate the amounts of reactants and products in a chemical reaction.

First, let's balance the equation:

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

The balanced equation indicates that 2 moles of AgCl are produced for every 1 mole of BaCl2. Therefore, we need to convert the given mass of BaCl2 into moles.

1. Calculate the molar mass of BaCl2:
Ba (barium) has a molar mass of 137.33 g/mol
Cl (chloride) has a molar mass of 35.45 g/mol

Molar mass of BaCl2 = (137.33 g/mol) + (2 x 35.45 g/mol) = 208.23 g/mol

2. Convert the given mass of BaCl2 into moles:
Moles of BaCl2 = (Given mass of BaCl2) / (Molar mass of BaCl2)
= 4.62 g / 208.23 g/mol
≈ 0.0222 mol

3. Use the mole ratio from the balanced equation to find the moles of AgCl:
Moles of AgCl = (Moles of BaCl2) x (2 mol AgCl / 1 mol BaCl2)
= 0.0222 mol x (2 mol AgCl / 1 mol BaCl2)
= 0.0444 mol

4. Convert the moles of AgCl into grams:
Mass of AgCl = (Moles of AgCl) x (Molar mass of AgCl)
= 0.0444 mol x (143.32 g/mol)
≈ 6.36 g

Therefore, approximately 6.36 grams of silver chloride can be produced from 4.62 grams of barium chloride.