Question

4) When 125 mL of 0.500 M AgNO3 is added to 100. mL of 0.500 M NH4CI, how many grams of AgCl are formed? (MM AgCI =143.32 g/mol) AgNO3(ag)+ NH4C| (aq) AgCl(s)+ NH4NO3(aq)

Answer 1

number of moles of AgNO3 = molarity * volume of solution in L

number of moles of AgNO3 = 0.500 * 0.125 = 0.0625 mole

Number of moles of NH4Cl = molarity * volume of solution in L

Number of moles of NH4Cl = 0.500 * 0.100 = 0.0500 mole

From the balanced equation we can say that

1 mole of AgNO3 requires 1 mole of NH4Cl so

0.0625 mole of AgNO3 will require

= 0.0625 mole of AgNO3 *(1 mole of NH4Cl / 1 mole of AgNO3)

= 0.0625 mole of NH4Cl

But we have 0.0500 mole of NH4Cl which is in short so NH4Cl is limiting reactant

From the balanced equation we can say that

1 mole of NH4Cl produces 1 mole of AgCl so

0.0500 mole of NH4Cl will produce

= 0.0500 mole of NH4Cl *(1 mole of AgCl / 1 mole of NH4Cl)

= 0.0500 mole of AgCl

mass of 1 mole of AgCl = 143.32 g so

the mass of 0.0500 mole of AgCl = 7.17 g

Therefore, the mass of AgCl produced would be 7.17 g