How many grams of silver chloride are produced from 5.0 g of silver nitrate reacting with an excess of barium chloride?
First, we need to determine the balanced chemical equation for the reaction between silver nitrate (AgNO3) and barium chloride (BaCl2):
2 AgNO3 + BaCl2 -> 2 AgCl + Ba(NO3)2
From the balanced equation, we can see that 2 moles of silver nitrate react with 1 mole of barium chloride to produce 2 moles of silver chloride.
Next, we need to determine the molar masses of the compounds involved:
AgNO3: 1 Ag (107.87 g/mol) + 1 N (14.01 g/mol) + 3 O (16.00 g/mol) = 169.88 g/mol
AgCl: 1 Ag (107.87 g/mol) + 1 Cl (35.45 g/mol) = 143.32 g/mol
Now we can determine the number of moles of silver nitrate in 5.0 g:
5.0 g / 169.88 g/mol = 0.0294 mol
Since 2 moles of silver nitrate react with 2 moles of silver chloride, we can use the mole ratio to determine the number of moles of silver chloride produced:
0.0294 mol AgNO3 x (2 mol AgCl / 2 mol AgNO3) = 0.0294 mol AgCl
Finally, we can calculate the mass of silver chloride produced using the molar mass of AgCl:
0.0294 mol AgCl x 143.32 g/mol = 4.22 g AgCl
Therefore, 5.0 g of silver nitrate would produce 4.22 grams of silver chloride when reacting with an excess of barium chloride.