When 31.2 mL of 0.500 mol L^-1 AgNO₃(aq) is added to 25.0 mL of 0.300 mol L^-1 NH₄Cl(aq) ,how many grams of AgCl are formed?
AgNO₃(aq) + NH₄Cl(aq) → AgCl(s) + NH₄NO₃(aq)

Question

Quizplus · Accepted Answer

First, we need to determine the limiting reactant. To do this, we can use the molarity and volume of each solution to calculate the number of moles of each reactant:

nAgNO3 = (0.500 mol/L)(0.0312 L) = 0.0156 mol
nNH4Cl = (0.300 mol/L)(0.0250 L) = 0.0075 mol

Since AgNO3 has more moles than NH4Cl, NH4Cl is the limiting reactant.

Next, we can use the stoichiometry of the balanced equation to calculate the number of moles of AgCl that should be formed:

1 mol NH4Cl produces 1 mol AgCl

nAgCl = nNH4Cl = 0.0075 mol

Finally, we can use the molar mass of AgCl to convert moles to grams:

mAgCl = nAgCl x MM AgCl
mAgCl = 0.0075 mol x 143.3 g/mol
mAgCl = 1.07 g

Therefore, the answer is (A) 1.07 g.

When 312 ML of 0