Question

show your calculation

When 100 mL of 0.250 M AgNO3 is added to 150. mL of 0.300 M NH4Cl, how many grams of AgCl are formed? AgNO3(aq) + NH4Cl(aq) → AgCl(s) + NH4NO3(aq) o 3.58 g O 8.96 g o 13.3 g 02.11g

Answer 1

Ag Cl (s) + NHy NO3 lag) * AgNO₃(aq) + NHy Q (aq) we have • loomL of 0.250 M AgNO3 ISOML of 0.300 M N Hyll Now, molnity Molarity = Number of Moles of solute . Volume of solution (inL) Number of moles of solute = Volume of solution X sor Molasity Number of moles of Ag NO₃ = 100 L X 0.250 ml 1000 = 0.025 mol - Number of moles of NHy U = 150 L x 0.300 ml 1000 = 0.045 mol This means that we hawl 0.025 mol of AgNO₃ and 0.045 mol of NHyu.

7 from the above balanced Chanical reaction it is clear that . I mole of AgNo reacts with Imole This means both deact in equal moles. But we have Moles of Ag NO3 < Moles of Nhu ce This means AgNO₃ will read totally before the Nhy ll reacts totally So, AgNO₃ is a too limiting reagents Now, I ole of AgNoz produces one mole of Ag q. -0.025 mol of AgNO3 will produce 0.025 mol of The mass of All = Number of moles X Molar Mass = 0.025 mol x 143.32 mol. = 3.5834

Answer : 3.58 g

Thank You.