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3 years ago

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What is the molarity of ZnCl2 that forms when 30.0 g of zinc completely reacts with CuCl2 according to the following reaction? A

ssume a final volume of 270 mL . Zn(s)+CuCl2(aq)→ZnCl2(aq)+Cu(s)

2 answers:

Alex73 [517]3 years ago

8 0

Answer:

The molarity of ZnCl2 is 1.699 mol/ L.

Explanation:

Balance chemical equation:

Zn(s) + CuCl2(aq) → ZnCl2(aq) + Cu(s)

Given data:

mass of zinc = 30 g

total volume of solution = 270 mL

molarity of ZnCl2 = ?

Solution:

First of all we will calculate the moles of Zn.

molar mass of zinc = 65.38 g/mol

number of moles = mass / molar mass

number of moles = 30 g/ 65.38 g/mol

number of moles = 0.4588 mole

From the balance chemical equation we will compare the moles of Zn and ZnCl2

Zn : ZnCl2

1 : 1

0.4588 : 0.4588

The number of moles of ZnCl2 are 0.4588 mol.

Molarity of ZnCl2:

Total volume of solution is 270 mL. We will convert it into liter.

1 L = 1000 mL

270/1000 = 0.27 L

Molarity = moles of solute / volume of solution in liter

Molarity = 0.4588 mol / 0.27 L

Molarity = 1.699 mol/ L

neonofarm [45]3 years ago

5 0

<u>Answer:</u>

The balanced chemical equation is

$Zn(s)+CuCl_2 (aq)>ZnCl_2 (aq)+Cu(s)$

The conversions are

Mass Zinc to moles Zinc (dividing by molar mass of Zinc)

Moles Zinc to moles $ZnCl_2$ ( using mole ratio 1 : 1 )

Moles $ZnCl_2$ to Molarity of $ZnCl_2$

$moles Zn= \frac {mass}{(molar mass)} \\\\=\frac {30.0g}{(65g per mol )} = 0.462 mol$

$0.462mol Zn \times \frac{(1mol ZnCl_2)}{(1mol Zn)}\\\\=0.462 mol ZnCl_2$

$Molarity ZnCl_2 = \frac {(moles ZnCl_2)}{(Volume of solution in L)}$

$= \frac {(0.462mol ZnCl_2)}{0.270L}=1.71 \frac {mol}{L} or M$

(Answer)

1.71 M is the molarity of $ZnCl_2$ that forms when 30.0 g of zinc completely reacts with $CuCl_2$

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Answer:

a) 6.12

b) 1.87

Explanation:

At the onset of the equivalence point (i.e the first equivalence point); alaninate is being converted to alanine.

$H_2NC_2H_5CO^-_2$ $+$ $H^+$ ------> $H_3}^+NC_2H_5CO^-_2$

1 mole of alaninate react with 1 mole of acid to give 1 mole of alanine;

therefore 50.0 mL of 0.160 M alaninate required 50.0 mL of 0.160M HCl to reach the first equivalence point.

The concentration of alanine can be gotten via the following process as shown below;

$[H_3}^+NC_2H_5CO^-_2]$ = $\frac{initial moles of alaninate}{total volume}$

$[H_3}^+NC_2H_5CO^-_2]$ = $\frac{(50.0mL)*(0.160M)}{(50.0mL+50.0mL)}$

$[H_3}^+NC_2H_5CO^-_2]$ = $\frac{8}{100mL}$

$[H_3}^+NC_2H_5CO^-_2]$ = 0.08 M

Alanine serves as an intermediary form, however the concentration of $H^+$ and the pH can be determined as follows;

$[H^+]$ = $\sqrt{\frac{K_{a1}K_{a2}{[H_3}^+NC_2H_5CO^-_2]+K_{a1}K_w}{ K_{a1}{[H_3}^+NC_2H_5CO^-_2] } }$

$[H^+]$ = $\sqrt{\frac{ (10^{-pK_{a1})}(10^{-pK_{a2})}(0.08)+(10^{-pK_{a1})}(1.0*10^{-14})} {(10^{-pK_{a1}})+(0.08)} }$

$[H^+]$ = $\sqrt{\frac{ (10^{-2.344})(10^{-9.868})(0.08)+(10^{-2.344})(1.0*10^{-14})} {(10^{-2.344})+(0.08)} }$

$[H^+]$ = $7.63*10^{-7}M$

pH = - log $[H^+]$

pH = $-log[7.63*10^{-7}]$

pH= 6.12

Therefore, the pH of the first equivalent point = 6.12

b) At the second equivalence point; all alaninate is converted into protonated alanine.

$H_2NC_2H_5CO^-_2$ $+$ $H^+$ -----> $H^+_3NC_2H_5CO^-_2$

$H^+_3NC_2H_5CO^-_2$ $+$ $H^+$ -----> $H^+_3NC_2H_5CO_2H$

Here; we have a situation where 1 mole of alaninate react with 2 moles of acid to give 1 mole of protonated alanine;

Moreover, 50.0 mL of 0.160 M alaninate is needed to produce 100.0mL of 0.160 M HCl in order to achieve the second equivalence point.

Thus, the concentration of protonated alanine can be determined as:

$[H^+_3NC_2H_5CO_2H]$ = $\frac{initial moles of alaninate}{total volume}$

$[H^+_3NC_2H_5CO_2H]$ = $\frac{(50.0mL)*(0.160M)}{(50.0mL+100.0mL)}$

$[H^+_3NC_2H_5CO_2H]$ = $\frac{8}{150}$

$[H^+_3NC_2H_5CO_2H]$ = 0.053 M

The pH at the second equivalence point can be calculated via the dissociation of protonated alanine at equilibrium which is represented as:

$H^+_3NC_2H_5CO_2H$ ⇄ $H^+_3NC_2H_5CO^-_2$ $+$ $H^+$

(0.053 - x) x x

$K_{a1}$ = $\frac{[H^+] [H^+_3NC_2H_5CO^-_2]}{[H^+_3NC_2H_5CO_2H]}$

$10^{-PK_{a1}}$ = $\frac{x*x}{(0.053-x)}$

$10^{-2.344} =\frac{x^2}{(0.053-x)}$

$0.00453 = \frac{x^2}{(0.053-x)}$

$0.00453(0.053-x) =x^2$

$x^2+0.00453x-(2.4009*10^{-4})$

Using quadratic equation formula;

$\frac{-b+/-\sqrt{b^2-4ac} }{2a}$

we have:

$\frac{-0.00453+\sqrt{(0.00453)^2-4(1)(-2.4009*10^{-4})} }{2(1)}$ OR $\frac{-0.00453-\sqrt{(0.00453)^2-4(1)(-2.4009*10^{-4})} }{2(1)}$

= 0.0134 OR -0.0179

So; we go by the positive integer which says

x = 0.0134

So $[H^+]=[H_3^+NC_2H_5CO^-_2]= 0.0134 M$

pH = $-log[H^+]$

pH = $-log[0.0134]$

pH = 1.87

Thus, the pH of the second equivalent point = 1.87

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