11. Silver ions can be precipitated from aqueous solutions by the addition of aqueous chloride: KCl(aq) + AgNO3(aq) → AgCl(s) + KNO3(aq)

Silver chloride is virtually insoluble in water so that the reaction appears to go to completion. How many grams of solid KCl must be added to 25.0 mL of 0.125 M AgNO3 solution to completely precipitate the silver?

0.2330 grams of solid KCl

Explanation:

In order to find KCl in grams, we need to convert from moles to moles to grams..

However, we are given the volume and the molarity, so first we will use that to convert to moles of AgNO3 (silver nitrate) and then use that to convert to moles of KCl and finally to grams of KCl..

-----------------------------------------------------------------------------------------------------------------

Step 1 -> We need to convert milliliters to liters

(25.0 ml)() = 0.025 L

Step 2 -> Convert from liters and molarity to just moles of AgNO3

(0.025 L)(0.125 )  =  0.003125  moles

**Liters cancel out, so just multiply across...

Step 3 -> Now that we have moles of AgNO3, we will convert to moles of KCl

We will do this using the balanced equation

We see that in the balanced equation, for every mol of AgNO3 that reacts, one mol of KCl will react too. So we will set the dimensional analysis up like the following:

(0.003125 moles AgNO3) *

Step 4-> So now that we have moles of KCl, we will just convert to grams

We will do this using the molar mass (atomic mass) of KCl to convert to grams

0.00312 mol KCl *

This can be rounded to 0.2330 grams of KCl

when 25.0 ml of 0.700 mol/l naoh was mixed in a calorimeter with 25.0 ml of 0.700 mol/l hcl, both initially at 20.0 °c, the temperature increased to 22.1 °c. the heat capacity of the calorimeter is 279 j/°c.

the equation for the reaction is:

naoh + hcl → nacl + h₂o

moles of hcl = 0.0250 l hcl ×

0.700

mol hcl

1 l hcl = 0.0175 mol hcl

volume of solution = (25.0 + 25.0) ml = 50.0 ml

mass of solution = 50.0 ml soln ×

1.00

ml soln

= 50.0 g soln

δt=t2-t1

= (22.1 – 20.0) °c = 2.1 °c

the heats involved are

heat from neutralization + heat to warm solution + heat to warm calorimeter = 0

q1+q2+q3=0

nδh+mcδt+cδt=0

0.0175 mol ×

δh

+ 50.0 g × 4.184 j·g⁻¹°c⁻¹ × 2.1 °c + 279 j°c⁻¹ × 2.1 °c = 0

0.0175 mol ×

δh

+ 439.32 j + 585.9 j = 0

0.0175 mol ×

δh

= -1025.22 j

δh=1025.22j0.0175 mol= -58 600 j/mol = -58.6 kj/mol