what volume in Liters of 0.15M KCl solution is required to completely react with 0.150 liter of a 0.175M Pb(NO3)2 solution according to the following balanced equation?
2KCl(aq)+ Pb(NO3)2(aq)-PbCl2(s)+ 2KNO3(aq)
moles Pb(NO3)2 = M x L = 0.175 x 0.150 = 0.02625.
From the equation, it takes 2 moles KCl to = 1 mole Pb(NO3)2; therefore, moles KCl = 2 x 0.02625 = 0.0525
Then M KCl = moles KCl/L KCl.
You know M and moles, solve for L.
To determine the volume of 0.15M KCl solution required to completely react with 0.150 liters of a 0.175M Pb(NO3)2 solution, we need to use the balanced equation and the concept of stoichiometry.
The balanced equation is as follows:
2KCl(aq) + Pb(NO3)2(aq) → PbCl2(s) + 2KNO3(aq)
From the equation, we can see that 2 moles of KCl react with 1 mole of Pb(NO3)2. This means that the mole ratio of KCl to Pb(NO3)2 is 2:1.
First, let's calculate the number of moles of Pb(NO3)2:
Moles of Pb(NO3)2 = concentration (M) × volume (L)
= 0.175 M × 0.150 L
= 0.02625 moles
Since the mole ratio of KCl to Pb(NO3)2 is 2:1, the number of moles of KCl required is also 0.02625 moles.
Now, we can calculate the volume of the 0.15M KCl solution needed.
Volume of KCl solution (L) = moles of KCl / concentration (M)
= 0.02625 moles / 0.15 M
≈ 0.175 L
Therefore, approximately 0.175 liters (or 175 mL) of 0.15M KCl solution is required to completely react with 0.150 liters of a 0.175M Pb(NO3)2 solution.