When 7.80 mL of 0.500 M AgNO₃ is added to 6.25 mL of 0.300 M NH₄Cl,how many grams of AgCl are formed?
AgNO₃(aq) + NH₄Cl(aq) → AgCl(s) + NH₄NO₃(aq)

Question

When 7.80 mL of 0.500 M AgNO₃ is added to 6.25 mL of 0.300 M NH₄Cl,how many grams of AgCl are formed? AgNO₃(aq)+ NH₄Cl(aq)→ AgCl(s)+ NH₄NO₃(aq) A) 0.269 g B) 0.553 g C) 0.822 g D) 1.61 g

Quizplus · Accepted Answer

The limiting reactant is NH₄Cl. Moles of NH₄Cl = 0.00625 L * 0.300 mol/L = 0.001875 mol. Moles of AgCl formed = 0.001875 mol. Mass of AgCl = 0.001875 mol * 143.32 g/mol = 0.269 g.

When 780 ML of 0