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givi [52]
2 years ago
6

Use your experimentally determined value of ksp and show,by calculations, that ag2cro4 should precipitate when 5ml of 0.004m agn

o3 are added to 5ml of 0.0024 m k2cro4.
Chemistry
1 answer:
Doss [256]2 years ago
8 0
When the value of Ksp = 3.83 x 10^-11 (should be given - missing in your Q)

So, according to the balanced equation of the reaction:

and by using ICE table:

              Ag2CrO4(s)  → 2Ag+ (Aq) + CrO4^2-(aq)

initial                                     0                   0

change                              +2X                 +X

Equ                                       2X                   X

∴ Ksp = [Ag+]^2[CrO42-]

so by substitution:

∴ 3.83 x 10^-11 = (2X)^2* X

3.83 x 10^-11 = 4 X^3

∴X = 2.1 x 10^-4 

∴[CrO42-] = X = 2.1 x 10^-4 M

[Ag+] = 2X = 2 * (2.1 x 10^-4) 

                  = 4.2 x 10^-4 M

when we comparing with the actual concentration of [Ag+] and [CrO42-]

when moles Ag+ = molarity * volume

                               = 0.004 m * 0.005L

                               = 2 x 10^-5 moles
[Ag+] = moles / total volume
     
          = 2 x 10^-5 / 0.01L

          = 0.002 M

moles CrO42- = molarity * volume

                         = 0.0024 m * 0.005 L

                         = 1.2 x 10^-5 mol

∴[CrO42-] = moles / total volume

                 = (1.2 x 10^-5)mol / 0.01 L 

                 = 0.0012 M

by comparing this values with the max concentration that is saturation in the solution 

and when the 2 values of ions concentration are >>> than the max values o the concentrations that are will be saturated.

∴ the excess will precipitate out       
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Answer:

Actual yield: 86.5 grams.

Explanation:

How many moles of formula units in 95 grams of calcium carbonate \rm CaCO_3?

Refer to a modern periodic table for relative atomic mass data:

  • Ca: 40.078;
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  • O: 15.999.

Formula mass of \rm CaCO_3:

M(\mathrm{CaCO_3})  = \underbrace{1\times 40.078}_{\rm Ca} + \underbrace{1\times 12.011}_{\rm C} + \underbrace{3\times 15.999}_{\rm O} = \rm 100.086\;g\cdot mol^{-1}.

\displaystyle n(\mathrm{CaCO_3}) = \frac{m(\mathrm{CaCO_3})}{M(\mathrm{CaCO_3})} = \rm \frac{95\;g}{100.086\;g\cdot mol^{-1}} = 0.949184\;mol.

How many moles of \rm CaCl_2 will be produced?

The coefficient in front of \rm CaCO_3 in the chemical equation is the same as that in front of \rm CaCl_2. That is:

\displaystyle \frac{n(\rm CaCl_2)}{n(\rm CaCO_3)} = 1.

\displaystyle n(\mathrm{CaCl_2}) = n(\mathrm{CaCO_3})\cdot \frac{n(\rm CaCl_2)}{n(\rm CaCO_3)} = n(\mathrm{CaCO_3}) = \rm 0.949184\;mol.

What's the theoretical yield of calcium chloride? In other words, what's the mass of \rm 0.949184\;mol of \rm CaCl_2?

Again, refer to a periodic table for relative atomic data:

  • Ca: 40.078;
  • Cl: 35.45.

M(\mathrm{CaCl_2}) = \underbrace{1\times 40.078}_{\rm Ca} + \underbrace{2\times 35.45}_{\rm Cl} = \rm 110.978\;g\cdot mol^{-1}.

\begin{aligned}m(\mathrm{CaCl_2}) &= n(\mathrm{CaCl_2})\cdot M(\mathrm{CaCl_2})\\ &= \rm 0.949184\;mol\times 110.978\;g\cdot mol^{-1}\\ &= \rm 105.339\; g\end{aligned}.

What's the actual yield of calcium chloride?

\displaystyle \text{Percentage Yield} = \frac{\text{Actual Yield}}{\text{Theoretical Yield}}\times 100\%.

\displaystyle \begin{aligned}\text{Actual Yield} &= \text{Theoretical Yield}\cdot \frac{\text{Percentage Yield}}{100\%}\\ &=\rm 105.339\; g \times \frac{82.15\%}{100\%}\\&= \rm 86.5\;g \end{aligned}.

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Answer:

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