How many grams of silver chloride can be produced if you start with 4.62 grams of barium chloride ? 2AgNO3 + BaCl2 —> 2AgCl + Ba(NO3)
2AgNO3 + BaCl2 —> 2AgCl + Ba(NO3)
mols BaCl2 = grams/molar mass = ?
Look at the coefficients in the balanced equation and note that 1 mol BaCl2 = 2 mols AgCl. Convert mols BaCl2 to mols AgCl.
Then convert mols AgCl to grams. grams = mols x molar mass = ?
To determine the amount of silver chloride that can be produced when starting with 4.62 grams of barium chloride, you need to calculate the molar mass of barium chloride and use stoichiometry.
1. Begin by finding the molar mass of BaCl2:
- The molar mass of Ba (barium) is 137.33 g/mol
- The molar mass of Cl (chlorine) is 35.45 g/mol (since there are two chlorine atoms, multiply this by 2)
- Add the molar masses together: 137.33 g/mol + (35.45 g/mol x 2) = 208.23 g/mol
2. Use the balanced chemical equation to determine the stoichiometric coefficients between barium chloride (BaCl2) and silver chloride (AgCl):
- From the equation: 2AgNO3 + BaCl2 -> 2AgCl + Ba(NO3)2
- The coefficient of barium chloride (BaCl2) is 1, and the coefficient of silver chloride (AgCl) is 2.
3. Apply stoichiometry to convert the mass of barium chloride to the mass of silver chloride using the molar ratios:
- Multiply the given mass of barium chloride (4.62 grams) by the molar mass ratio between the two compounds:
4.62 g BaCl2 x (2 mol AgCl / 1 mol BaCl2) x (143.32 g AgCl / 1 mol AgCl) = 13.14 g AgCl
Therefore, starting with 4.62 grams of barium chloride, you can produce approximately 13.14 grams of silver chloride.