How many grams of silver chloride are produced from 5.0 g of silver nitrate reacting with an excess of barium chloride?
How much barium chloride is necessary to react with the silver nitrate
To determine the amount of silver chloride produced from 5.0 g of silver nitrate, we need to first write out the balanced chemical equation for the reaction:
2AgNO3 + BaCl2 -> 2AgCl + Ba(NO3)2
From this equation, we can see that 2 moles of silver nitrate react with 1 mole of barium chloride to produce 2 moles of silver chloride.
First, we need to calculate the number of moles of silver nitrate involved in the reaction:
moles of AgNO3 = mass / molar mass
moles of AgNO3 = 5.0 g / 169.87 g/mol = 0.0294 mol
Since the reaction involves a 2:1 ratio of silver nitrate to silver chloride, the number of moles of silver chloride produced will also be 0.0294 mol.
Now, we can calculate the mass of silver chloride produced:
mass of AgCl = moles * molar mass
mass of AgCl = 0.0294 mol * 143.32 g/mol = 4.21 g
Therefore, 4.21 grams of silver chloride are produced from 5.0 grams of silver nitrate reacting with an excess of barium chloride.
To determine the amount of barium chloride necessary to react with the silver nitrate, we first need to calculate the number of moles of silver nitrate involved in the reaction (as calculated above):
moles of AgNO3 = 0.0294 mol
Since the reaction involves a 1:2 ratio of silver nitrate to barium chloride, the number of moles of barium chloride needed will be 2 times the number of moles of silver nitrate:
moles of BaCl2 = 2 * moles of AgNO3
moles of BaCl2 = 2 * 0.0294 mol = 0.0588 mol
Finally, we can calculate the mass of barium chloride needed:
mass of BaCl2 = moles * molar mass
mass of BaCl2 = 0.0588 mol * 208.23 g/mol = 12.26 g
Therefore, 12.26 grams of barium chloride are necessary to react with 5.0 grams of silver nitrate.