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e-lub [12.9K]
3 years ago
11

The following data was collected when a reaction was performed experimentally in the laboratory.

Chemistry
1 answer:
REY [17]3 years ago
5 0

Answer:

The answer to your question is 3 moles of AlCl₃

Explanation:

Process

1.- Write and balance the equation

                  Al(NO₃)₃ + 3NaCl   ⇒   3NaNO₃  +  AlCl₃

2.- Determine the limiting reactant

Theoretical proportion     1 mol Al(NO₃)₃ :  3 moles of NaCl                  

Experimental proportion   4 moles Al(NO₃)₃ : 9 moles NaCl

From these values, we determine that the limiting reactant is NaCl because the number of moles increases three times and the number of moles of Al(NO₃)₃  increases four times.

3.- Determine the amount of AlCl₃ using proportions

                       3 moles of NaCl --------------- 1 mol of AlCl₃

                       9 moles of NaCl ----------------  x

                       x = (9 x 1) / 3

                       x = 9 /3

                       x = 3 moles

                     

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Calculate the solubility at 25°C of CuBr in pure water and in a 0.0120M CoBr2 solution. You'll find Ksp data in the ALEKS Data t
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Answer:

S = 7.9 × 10⁻⁵ M

S' = 2.6 × 10⁻⁷ M

Explanation:

To calculate the solubility of CuBr in pure water (S) we will use an ICE Chart. We identify 3 stages (Initial-Change-Equilibrium) and complete each row with the concentration or change in concentration. Let's consider the solution of CuBr.

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<u>Solubility in 0.0120 M CoBr₂ (S')</u>

First, we will consider the ionization of CoBr₂, a strong electrolyte.

CoBr₂(aq) → Co²⁺(aq) + 2 Br⁻(aq)

1 mole of CoBr₂ produces 2 moles of Br⁻. Then, the concentration of Br⁻ will be 2 × 0.0120 M = 0.0240 M.

Then,

    CuBr(s) ⇄ Cu⁺(aq) + Br⁻(aq)

I                       0           0.0240

C                     +S'           +S'

E                       S'            0.0240 + S'

Ksp = 6.27 × 10⁻⁹ = [Cu⁺].[Br⁻] = S' . (0.0240 + S')

In the term (0.0240 + S'), S' is very small so we can neglect it to simplify the calculations.

S' = 2.6 × 10⁻⁷ M

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