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riadik2000 [5.3K]
1 year ago
15

7.0 mol Mn reacts with 5.0 mol

Chemistry
2 answers:
amm18121 year ago
7 0

Answer:

5.0 mol

Explanation:

From the coefficients of the equation, we know that for every 2 moles of Mn consumed, 2 moles of MnO are produced.

Thus, the number of moles of Mn consumed is equal to the number of moles of MnO produced, and thus the answer is 5.0 mol.

Elenna [48]1 year ago
6 0

The moles of MnO formed with the reaction of 5 moles of Mn is 3 moles.

<h3>What is a limiting reagent?</h3><h3 />

In a chemical reaction, the reagent that is present in a lesser quantity and governs the rate of the reaction is termed as the limiting reagent.

In the reaction of the formation of MnO, according to the stoichiometric law 2 moles of Mn reacts with 1 moles of Oxygen.

Thus, the moles of oxygen consumed by 7 moles of Mn is:

2 moles Mn = 1 mole O₂

7 moles Mn = 3.5 moles O₂

The available moles of O₂ = 5 mol.

The remaining moles of O₂ = 5-3.5 moles

The remaining moles of O₂ = 1.5 moles

The reaction of 5 moles Mn requires 2.5 moles of O₂, whereas the available moles of oxygen is 1.5 moles. Thus, oxygen serves as the limiting reagent.

The moles of MnO formed with the reaction of 5 moles Mn and 1.5 moles O₂

1 mole O₂ = 2 moles MnO

1.5 moles O₂ = 2 * 1.5 moles MnO

1.5 moles O₂ = 3 moles MnO

Thus, the moles of MnO formed with the reaction of 5 moles of Mn with the available oxygen is 3 moles.

Learn more about stoichiometric law, here:

brainly.com/question/14465605

#SPJ1

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A 27.9 mL sample of 0.289 M dimethylamine, (CH3)2NH, is titrated with 0.286 M hydrobromic acid.
sesenic [268]

Answer:

(1) Before the addition of any HBr, the pH is 12.02

(2) After adding 12.0 mL of HBr, the pH is 10.86

(3) At the titration midpoint, the pH is 10.73

(4) At the equivalence point, the pH is 5.79

(5) After adding 45.1 mL of HBr, the pH is 1.18

Explanation:

First of all, we have a weak base:

  • 0 mL of HBr is added

(CH₃)₂NH  + H₂O  ⇄  (CH₃)₂NH₂⁺  +  OH⁻            Kb = 5.4×10⁻⁴

0.289 - x                             x                x

Kb = x² / 0.289-x

Kb . 0.289 - Kbx - x²

1.56×10⁻⁴ - 5.4×10⁻⁴x - x²

After the quadratic equation is solved x = 0.01222 → [OH⁻]

- log  [OH⁻] = pOH → 1.91

pH = 12.02   (14 - pOH)

  • After adding 12 mL of HBr

We determine the mmoles of H⁺, we add:

0.286 M . 12 mL = 3.432 mmol

We determine the mmoles of base⁻, we have

27.9 mL . 0.289 M = 8.0631 mmol

When the base, react to the protons, we have the protonated base plus water (neutralization reaction)

(CH₃)₂NH     +      H₃O⁺        ⇄  (CH₃)₂NH₂⁺  +  H₂O

8.0631 mm       3.432 mm                 -

4.6311 mm                                  3.432 mm

We substract to the dimethylamine mmoles, the protons which are the same amount of protonated base.

[(CH₃)₂NH] → 4.6311 mm / Total volume (27.9 mL + 12 mL) = 0.116 M

[(CH₃)₂NH₂⁺] → 3.432 mm / 39.9 mL = 0.0860 M

We have just made a buffer.

pH = pKa + log (CH₃)₂NH  / (CH₃)₂NH₂⁺

pH = 10.73 + log (0.116/0.0860) = 10.86

  • Equivalence point

mmoles of base = mmoles of acid

Let's find out the volume

0.289 M . 27.9 mL = 0.286 M . volume

volume in Eq. point = 28.2 mL

(CH₃)₂NH     +      H₃O⁺        ⇄  (CH₃)₂NH₂⁺  +  H₂O

8.0631 mm       8.0631mm               -

                                                8.0631 mm

We do not have base and protons, we only have the conjugate acid

We calculate the new concentration:

mmoles of conjugated acid / Total volume (initial + eq. point)

[(CH₃)₂NH₂⁺] = 8.0631 mm /(27.9 mL + 28.2 mL)  = 0.144 M

(CH₃)₂NH₂⁺   +  H₂O   ⇄   (CH₃)₂NH  +  H₃O⁻       Ka = 1.85×10⁻¹¹

 0.144 - x                                  x               x

[H₃O⁺] = √ (Ka . 0.144) →  1.63×10⁻⁶ M  

pH = - log [H₃O⁺] = 5.79

  • Titration midpoint (28.2 mL/2)

This is the point where we add, the half of acid. (14.1 mL)

This is still a buffer area.

mmoles of H₃O⁺ = 4.0326 mmol (0.286M . 14.1mL)

mmoles of base = 8.0631 mmol - 4.0326 mmol

[(CH₃)₂NH] = 4.0305 mm / (27.9 mL + 14.1 mL) = 0.096 M

[(CH₃)₂NH₂⁺] = 4.0326 mm (27.9 mL + 14.1 mL) = 0.096 M

pH = pKa + log (0.096M / 0.096 M)

pH = 10.73 + log 1 =  10.73

Both concentrations are the same, so pH = pKa. This is the  maximum buffering capacity.

  • When we add 45.1 mL of HBr

mmoles of acid = 45.1 mL . 0.286 M = 12.8986 mmol

mmoles of base = 8.0631 mmoles

This is an excess of H⁺, so, the new [H⁺] = 12.8986 - 8.0631 / Total vol.

(CH₃)₂NH     +      H₃O⁺        ⇄  (CH₃)₂NH₂⁺  +  H₂O

8.0631 mm     12.8986 mm             -

       -               4.8355 mm                        

[H₃O⁺] = 4.8355 mm / (27.9 ml + 45.1 ml)

[H₃O⁺] = 4.8355 mm / 73 mL → 0.0662 M

- log [H₃O⁺] = pH

- log 0.0662 = 1.18 → pH

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Determine the ionic strength, μ, for each of the solutions. Assume complete dissociation of each salt and ignore any hydrolysis
kow [346]

Answer:

The answers are: (a )0.0058 M; (b) 0.00681 M; (c) 0.006912 M

Explanation:

Ionic strenght= μ= 1/2 ∑ c z²

Where c is the concentration of each ion (in M) and z is the charge of the ion. So, to calculate the ionic strenght of a solution you have to know the concentration of each ion and their charges. For this, the dissociation equilibrium is required.

a) HCl →    H⁺    +    Cl⁻

Conc H⁺= 0.00580 M

Conc Cl⁻= 0.00580 M

μ= 1/2 x ((Conc H⁺)(+1)²) + ((Conc Cl⁻)(-1)²

μ= 1/2 x ((0.00580 M) + (0.00580 M))

μ= 0.00580 M

b) CaBr₂  →  Ca²⁺ + 2 Br⁻

Conc Ca²⁺= 0.00227 M

Conc Br⁻= 2 x 0.00227 M= 0.00454 M

μ= 1/2 ((Conc Ca²⁺)(+2)² + ((Conc Br⁻)(-1)²)

μ= 1/2 (0.00227 M (4)) + (0.00454 M)

μ= 0.00681 M

c) The solution has two dissociation equilibria:

Mg(NO₃)₂ → Mg²⁺ + 2 NO₃⁻

La(NO₃)₃ → La³⁺ + 3 NO₃⁻

Conc Mg²⁺= 0.000752 M

Conc NO₃⁻= 2 x (0.000752 M) + 3 x (0.000776 M)= 0.003832 M

Conc  La³⁺= 0.000776 M

μ= 1/2 ((Conc Mg²⁺)(+2)²)+ ((Conc NO₃⁻)(-1)²) + ((Conc  La³⁺)(+3)²)

μ= 1/2 (0.000752 M(4)) + (0.003832 M) + (0.000776 M)(9))

μ= 0.006912 M

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