What mass of MnCl2 is produced when 0.091 g of Cl2 is generated?

Answers

Answer 1
MMn = 54.94g/mol
MO2 = 2(16) = 32g/mol
MH = 1g/mol
MCl = 35.45g/mol
Molar Mass of MnO2:
54.94
+
2
(
16
)
=
86.94
Molar Mass of HCl:
1
+
35.45
=
36.45
Looking at the equation, we need a 4:1 ratio of HCl to MnO2 to make the reaction work, let's see if we have four times as much of the HCl.
Mols of MnO2:
42.7
86.94
=
0.49
Mols of HCl:
47.1
36.45
=
1.29
As we can see, we have less than the amount of HCl we need for all of the MnO2 to react, therefore the HCl is the limiting reactant.
For the theoretical yield of Cl2, we need to determine how many mols will be produced. As we can see by the equation, four mols of HCl make one mol of Cl2.
Molar Mass of Cl2:
35.45

2
=
70.9
Since we have 1.29mol of HCl, and we know it is the limiting reactant:
Mols of Cl2
1.29
4
=
0.323
Mass of Cl2 (Theoretical yield)
0.323

70.9
=
22.9
To calculate the actual yield, we multiply the theoretical yield by the final percentage:
22.9

0.791
=
18.11
Therefore, the actual yield of chlorine is 18.11
Answer 2

The mass of MnCl2 that is produced when 0.091 g of Cl2 is generated is 0.16 g.

Molar mass of MnCl₂ = 55 + 35.5(2) = 126 g/mol

Molar mass of Cl₂ = 35.5(2) = 71 g/mol

If 71 g of Cl₂ gives 126 g of MnCl₂

0.091 g of Cl₂ will give ? of MnCl₂

cross multiply

mass of MnCl₂  produced  = 0.091 × 126 / 71

mass of MnCl₂  produced  = 11.466 / 71

mass of MnCl₂  produced  = 0.16149295774

mass of MnCl₂  produced  ≈ 0.16 g

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Answers

Answer:

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Answer:

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correct on edge2021

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Answers

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Answers

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Answers

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Answer:

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Answers

Answer:

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Math

Pre-Algebra

Order of Operations: BPEMDAS

BracketsParenthesisExponentsMultiplicationDivisionAdditionSubtractionLeft to Right

Chemistry

Atomic Structure

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Stoichiometry

Using Dimensional AnalysisExplanation:

Step 1: Define

280 g Cl₂

Step 2: Identify Conversions

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Step 3: Convert

[DA] Set up:                                                                                                   [tex]\displaystyle 280 \ g \ Cl_2(\frac{1 \ mol \ Cl_2}{79.90 \ g \ Cl_2})(\frac{22.4 \ L \ Cl_2}{1 \ mol \ Cl_2})[/tex][DA] Multiply/Divide [Cancel out units]:                                                       [tex]\displaystyle 78.4981 \ L \ Cl_2[/tex]

Step 4: Check

Follow sig fig rules and round. We are given 2 sig figs.

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Answers

Answer:

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step by step solution

Answers

Answer:

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Answers

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Answer:

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Answers

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Answers

Answer

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Answers

Answer:

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