When 15.6 mL of 0.500 M AgNO₃ is added to 12.5 mL of 0.300 M NH₄Cl,how many grams of AgCl are formed? AgNO₃(aq)+ NH₄Cl(aq)→ AgCl(s)+ NH₄NO₃(aq) A) 0.537 g B) 1.11 g C) 1.65 g D) 3.22 g

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Quizplus · Accepted Answer

First, correct the chemical formulas: AgNO3 + NH4Cl → AgCl + NH4NO3. To find the mass of AgCl formed, first determine the limiting reactant. Moles of AgNO3 = 15.6 mL * 0.500 M = 0.0078 mol. Moles of NH4Cl = 12.5 mL * 0.300 M = 0.00375 mol. NH4Cl is the limiting reactant. Moles of AgCl formed = moles of NH4Cl = 0.00375 mol. Mass of AgCl = moles * molar mass = 0.00375 mol * 143.32 g/mol = 0.537 g.

When 156 ML of 0