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alexandr402 [8]
3 years ago
8

Do the problem and show how you found the answer. Thanks!

Chemistry
1 answer:
patriot [66]3 years ago
3 0

Answer:

60.6

Explanation:

Given data:

Mass of Si = 10 g

Mass of Cl₂ = 60.0 g

Mass of SiCl₄ = ?

Solution:

Chemical equation:

Si + 2Cl₂  →   SiCl₄

Number of moles of Si:

Number of moles = mass/ molar mass

Number of moles = 10 g/28 g/mol

Number of moles = 0.357 mol

Number of moles of Cl₂ :

Number of moles = mass/ molar mass

Number of moles = 60 g/70.9 g/mol

Number of moles = 0.846 mol

Now we will compare the moles of SiCl₄ with both reactants.

                               Cl₂              :            SiCl₄

                                2                :              1

                             0.846           :           1/2×0.846=0.423

                                Si               :           SiCl₄

                                 1                 :            1

                                0.357         ;          0.357

Number of moles of SiCl₄ produce by Si are less so it will limiting reactant and limit the yield of SiCl₄.

Mass of SiCl₄:

Mass = number of moles × molar mass

Mass = 0.357 mol  × 169.8 g/mol

Mass = 60.6 g

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What is the pH of a buffer solution upon mixing 15.0 mL of 0.40 M HCl and 20.0 mL of 0.50 M NH? Kb (NH3) = 1.8 x 10 E. 7.00 A. 9
vladimir2022 [97]

<u>Answer:</u> The pH of resulting solution is 9.08

<u>Explanation:</u>

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

\text{Molarity of the solution}=\frac{\text{Moles of solute}\times 1000}{\text{Volume of solution (in mL)}}       ........(1)

  • <u>For HCl:</u>

Molarity of HCl = 0.40 M

Volume of solution = 15.0 mL

Putting values in equation 1, we get:

0.40M=\frac{\text{Moles of HCl}\times 1000}{15.0mL}\\\\\text{Moles of HCl}=0.006mol

  • <u>For ammonia:</u>

Molarity of ammonia = 0.50 M

Volume of solution = 20.0 mL

Putting values in equation 1, we get:

0.50M=\frac{\text{Moles of ammonia}\times 1000}{20.0mL}\\\\\text{Moles of ammonia}=0.01mol

The chemical reaction for hydrochloric acid and ammonia follows the equation:

                  HCl+NH_3\rightarrow NH_4Cl

Initial:          0.006      0.01

Final:             -         0.004              0.006

Volume of solution = 15.0 + 20.0 = 35.0 mL = 0.035 L    (Conversion factor:  1 L = 1000 mL)

  • To calculate the pOH of basic buffer, we use the equation given by Henderson Hasselbalch:

pOH=pK_b+\log(\frac{[salt]}{[base]})

pOH=pK_b+\log(\frac{[NH_4Cl]}{[NH_3]})

We are given:

pK_b = negative logarithm of base dissociation constant of ammonia = -\log (1.8\times 10^{-5})=4.74

[NH_4Cl]=\frac{0.006}{0.035}

[NH_3]=\frac{0.004}{0.035}

pOH = ?

Putting values in above equation, we get:

pOH=4.74+\log(\frac{0.006/0.035}{0.004/0.035})\\\\pOH=4.92

To calculate pH of the solution, we use the equation:

pH+pOH=14\\pH=14-4.92=9.08

Hence, the pH of the solution is 9.08

3 0
4 years ago
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