Answer:
4.21 g of AgCl
3.06 g of BaCl₂ will be needed to complete the reaction
Explanation:
The first step is to determine the reaction.
Reactants: BaCl₂ and AgNO₃
The products will be the silver chloride (AgCl) and the Ba(NO₃)₂
The reaction is: BaCl₂(aq) + 2AgNO₃(aq) → 2AgCl(s) ↓ + Ba(NO₃)₂ (aq)
We determine the silver nitrate moles: 5 g . 1mol / 169.87 g = 0.0294 moles. Now, according to stoichiometry, we know that ratio is 2:2-
2 moles of nitrate can produce 2 moles of chloride, so the 0.0294 moles of silver nitrate, will produce the same amount of chloride.
We convert the moles to mass → 143.32 g / mol . 0.0294 mol = 4.21 g of AgCl.
Now, we consider the BaCl₂.
2 moles of nitrate can react to 1 mol of barium chloride
Then, 0.0294 moles of silver nitrate will react to (0.0294 . 1) /2 = 0.0147 moles. We convert the moles to mass:
0.0147 mol . 208.23 g /1mol = 3.06 g of BaCl₂
