When 31.2 mL of 0.500 M AgNO₃ is added to 25.0 mL of 0.300 M NH₄Cl,how many grams of AgCl are formed?
AgNO₃(aq) + NH₄Cl(aq) → AgCl(s) + NH₄NO₃(aq)

Question

When 31.2 mL of 0.500 M AgNO₃ is added to 25.0 mL of 0.300 M NH₄Cl,how many grams of AgCl are formed? AgNO₃(aq)+ NH₄Cl(aq)→ AgCl(s)+ NH₄NO₃(aq)

Quizplus · Accepted Answer

The balanced chemical equation is:AgNO3(aq) + NH4Cl(aq) → AgCl(s) + NH4NO3(aq)From the stoichiometry of the balanced equation, we can see that 1 mole of AgNO3 reacts with 1 mole of NH4Cl to produce 1 mole of AgCl.Therefore, the number of moles of AgCl formed is equal to the number of moles of AgNO3 or NH4Cl used, whichever is limiting.To determine the limiting reactant, we can calculate the number of moles of each reactant:moles of AgNO3 = 0.500 M * 0.0312 L = 0.0156 molmoles of NH4Cl = 0.300 M * 0.0250 L = 0.00750 molSince the number of moles of NH4Cl is less than the number of moles of AgNO3, NH4Cl is the limiting reactant.Therefore, the number of moles of AgCl formed is equal to the number of moles of NH4Cl used:moles of AgCl = 0.00750 molThe molar mass of AgCl is 143.32 g/mol. Therefore, the mass of AgCl formed is:mass of AgCl = 0.00750 mol * 143.32 g/mol = 1.07 gTherefore, the correct answer is A) 1.07 g.

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