Answer:
Step-by-step explanation:
To determine the amount of barium chloride necessary to react with 35.0 g of silver nitrate, you need to use the balanced chemical equation for the reaction:
2AgNO3 + BaCl2 -> 2AgCl + Ba(NO3)2
This equation tells us that for every 2 moles of silver nitrate (AgNO3), 1 mole of barium chloride (BaCl2) is needed. To convert the number of grams of silver nitrate to moles, you can use its molar mass:
AgNO3:Ag = 107.87 g/mol
35.0 g of AgNO3 is equivalent to 35.0 g / 107.87 g/mol = 0.3230 moles of AgNO3
Since 1 mole of BaCl2 is needed for every 2 moles of AgNO3, we can calculate the number of moles of BaCl2 required:
0.3230 moles / 2 = 0.1615 moles of BaCl2
Finally, to convert moles of BaCl2 to grams, use its molar mass:
BaCl2:Ba = 208.23 g/mol
0.1615 moles * 208.23 g/mol = 33.45 g of BaCl2
So, to react with 35.0 g of silver nitrate, you would need 33.45 g of barium chloride.