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Jet001 [13]
2 years ago
6

How many grams of C12 can be prepared from the reaction of 16.0 g of MnO2 and 30.0 g of HCl according to the following chemical

equation? Mn02 + 4HCI → MnCl2 + Cl2 + 2H2O

Chemistry
1 answer:
kogti [31]2 years ago
8 0

Answer:

Mass of Cl₂ produced 12.78 g

Explanation:

Given data:

Mass of MnO₂ = 16 g

Mass of HCl = 30.0 g

Mass of Cl₂ produced = ?

Solution:

Chemical equation:

MnO₂ + 4HCl        →        MnCl₂ + Cl₂ + 2H₂O

Number of moles of MnO₂:

Number of moles = mass / molar mass

Number of moles =  16 g/ 87 g/mol

Number of moles = 0.18 mol

Number of moles of HCl:

Number of moles = mass / molar mass

Number of moles =  30 g/ 36.46 g/mol

Number of moles = 0.82 mol

Now we will compare the moles of Cl₂  with MnO₂ and HCl.

                    MnO₂               :                Cl₂

                        1                   :                   1

                       0.18               :                0.18

                    HCl                  :                   Cl₂  

                       4                    :                     1

                      0.82               :                 1/4×0.82 = 0.205 mol

The number of moles of Cl₂ formed by HCl are less it will limiting reactant.

Mass of Cl₂ formed:

Mass = number of moles × molar mass

Mass = 0.18 mol × 71 g/mol

Mass = 12.78 g

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8 0
2 years ago
How many grams of Boron can be obtained from 234 grams of B2O3?
Xelga [282]

Answer:

72.67g of B

Explanation:

The reaction of B₂O₃ to produce boron (B), is:

B₂O₃ → 3/2O₂ + 2B

<em>That means B₂O₃ produce 2 moles of boron</em>

Molar mass of B₂O₃ is 69.62g/mol. 234g of B₂O₃ contains:

234g B₂O₃ ₓ (1mol / 69.62g) = 3.361 moles of B₂O₃.

As 1 mole of B₂O₃ produce 2 moles of B, Moles of B that can be produced from B₂O₃ is:

3.361mol B₂O₃ ₓ 2 = <em>6.722 moles of B</em>.

As molar mass of B is 10.811g/mol. Thus mass of B that can be produced is:

6.722mol B ₓ (10.811g / mol) = <em>72.67g of B</em>

4 0
2 years ago
When the reaction mixture is worked-up, it is first washed three times with 5% sodium bicarbonate, and then with a saturated nac
Ann [662]

Solution:

After the reaction of mixture is worked-up Washing three times the organic  with sodium carbonate helps to decrease the solubility of the organic layer into the aqueous layer. This allows the organic layer to be separated more easily.

And then the reaction washed by saturated NACL we have The bulk of the water can often be removed by shaking or "washing" the organic layer with saturated aqueous sodium chloride (otherwise known as brine). The salt water works to pull the water from the organic layer to the water layer.

5 0
2 years ago
A 600.0 mL sample of 0.20 MHF is titrated with 0.10 MNaOH. Determine the pH of the solution after the addition of 600.0 mL of Na
Leona [35]

Answer: pH=12.69

Explanation:

{\text{Moles of HF}=Molarity\times {\text{Volume of solution in liters}}

{\text{Moles of HF}=0.20M\times 0.6L=0.12 moles

HF\rightarrow H^++F^-

Initial 0.12               0       0

Eqm   0.12-x           x        x

K_a=\frac{[H^+][F^-]}{[HF]}

3.5\times 10^{-4}=\frac{x^2}{0.12-x}  

(neglecting small value of x in comparison to 0.12)

x=4.2\times 10^{-5}

Moles of H^+=4.2\times 10^{-5}

NaOH\rightarrow Na^++OH^-

{\text{Molesof NaOH}}=Molarity\times {\text{Volume of solution in liters}}

{\text{Moles of NaOH}}=0.10M\times 0.6L=0.06 moles

0.06 moles of NaOH will give 0.06 moles of [OH^-]

Now 4.2\times 10^{-5} moles of OH^- will be neutralized by 4.2\times 10^{-5} moles of H^+ and (0.06-4.2\times 10^{-5})=0.059 moles of OH^- will be left.

Molarity of OH^-=\frac{0.059moles}{1.2L}=0.049M

pOH=-\log[OH^-]=-\log[0.049]=1.31

pH = 14 - pOH= 14 - 1.31 = 12.69

5 0
2 years ago
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Kaylis [27]

Physical changes occur when the properties of a substance are retained and/or the materials can be recovered after the change. Chemical changes involve the formation of a new substance. Formation of a gas, solid, light, or heat are possible evidence of chemical change.

6 0
2 years ago
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