Question

When 31.2 mL of 0.500 M AgNO₃ is added to 25.0 mL of 0.300 M NH₄Cl, how many grams AgCl are formed?

AgNO₃(aq) + NH₄Cl(aq) → AgCl(s)+ NH₄NO₃(aq)
A) 3.31 g
B) 2.24 g
C) 1.07g
D) 6.44 g

Answer 1

Therefore, the correct answer is B) 2.24 g.To find the amount of AgCl formed, we calculated the moles of each reactant and identified AgNO₃ as the limiting reactant.

Step-by-step explanation:

To calculate the amount of AgCl precipitate formed when 31.2 mL of 0.500 M AgNO₃ is mixed with 25.0 mL of 0.300 M NH₄Cl, we need to use a stoichiometric calculation based on the balanced chemical equation: AgNO₃(aq) + NH₄Cl(aq) → AgCl(s) + NH₄NO₃(aq).

First, we find the moles of AgNO₃ and NH₄Cl:

• Moles of AgNO₃ = Volume (L) × Molarity (mol/L) = 0.0312 L × 0.500 mol/L = 0.0156 mol
• Moles of NH₄Cl = 0.025 L × 0.300 mol/L = 0.0075 mol

Since the reaction is 1:1, AgNO₃ is the limiting reactant. So, all 0.0156 mol of AgNO₃ will react to form an equivalent amount of AgCl.

Finally, we convert the moles of AgCl to grams:

• Mass of AgCl = Moles × Molar Mass = 0.0156 mol × 143.32 g/mol ≈ 2.24 g
• Using stoichiometry, we then converted the moles of AgCl to grams, resulting in 2.24 g of AgCl being formed.