When 125 mL of 0.500 M AgNO₃ is added to 100.mL of 0.300 M NH₄Cl,how many grams of AgCl are formed? AgNO₃(aq)+ NH₄Cl(aq)→ AgCl(s)+ NH₄NO₃(aq) A)4.30 g B)8.96 g C)13.3 g D)25.8 g

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Quizplus · Accepted Answer

First, we need to identify the limiting reactant. The moles of AgNO3 are calculated as $0.125 \, \text{L} \times 0.500 \, \text{M} = 0.0625 \, \text{moles}$, and the moles of NH4Cl are $0.100 \, \text{L} \times 0.300 \, \text{M} = 0.0300 \, \text{moles}$. Since the reaction is 1:1, NH4Cl is the limiting reactant. The moles of AgCl formed will be equal to the moles of NH4Cl, which is 0.0300 moles. The molar mass of AgCl is approximately 143.32 g/mol, so the mass of AgCl formed is $0.0300 \, \text{moles} \times 143.32 \, \text{g/mol} = 4.30 \, \text{g}$.

When 125 ML of 0