Answer:
The answer to your question is 0.269 grams of AgCl
Explanation:
Data
[AgNO₃] = 0.50 M
Vol AgNO₃ = 7.80 ml
[NH₄Cl] = 0.30 M
Vol NH₄Cl = 6.25 ml
mass of AgCL
Balanced reaction
AgNO₃(aq) + NH₄Cl(aq) ⇒ AgCl (s) + NH₄NO₃ (aq)
Process
1.- Calculate the moles of AgNO₃
Molarity = moles / volume
moles = Molarity x volume
moles = 0.50 x 0.0078
moles = 0.0039
2.- Calculate the moles of NH₄Cl
moles = 0.30 x 0.0063
moles = 0.00188
3.- Calculate the limiting reactant
The proportion of AgNO₃(aq) to NH₄Cl(aq) is 1 :1, then, we conclude that the limiting reactant is NH₄Cl(aq), because there are less amount of this reactant in the experiment.
4.- Calculate the moles of AgCl
1 mol of NH₄Cl ---------------- 1 mol of AgCl
0.00188 mol of NH₄Cl ------------- x
x = (0.00188 x 1) /1
x = 0.00188 moles of AgCl
5.- Calculate the grams of AgCl
molecular mass of AgCl = 108 + 35.5 = 143.5 g
143.5 grams of AgCl -------------- 1 mol
x -------------------------------------------0.00188 moles of AgCl
x = (0.00188 x 143.5) / 1
x = 0.269 grams of AgCl
