Consider the following unbalanced equation: agno3 (aq) cacl2 (aq) → agcl (s) ca(no3)2 (aq) how many grams of silver chloride can be prepared by the reaction of 100.0 ml of 0.20 m silver nitrate with 100.0 ml of 0.15 m calcium chloride?

3 grams

Explanation:-

To calculate the number of moles for given molarity, we use the equation:

    .....(1)

Molarity of solution = 0.20 M

Volume of solution = 100.0 mL = 0.1 L

Putting values in equation 1, we get:

Molarity of solution = 0.15 M

Volume of solution = 100.0 mL = 0.1 L

Putting values in equation 1, we get:

The balanced chemical equation for the reaction is:

According to stoichiometry :

2 moles of require 1 mole of

Thus 0.02 moles of will require=  of

Thus is the limiting reagent as it limits the formation of product and is the excess reagent.

As 2 mole of give = 2 mole of

Thus 0.02 moles of give =  of

Mass of

Thus 3 g of will be produced from the given masses of both reactants.

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

i think the answer is terrae but not 100 percent sure. ▓▬▬ §