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3 years ago

How many milliliters of 0.021 moles of oxalic acid is needed to react with 80 mL of 0.11 moles of KMnO4

Chemistry

1 answer:

alexira [117]3 years ago

8 0

Answer:

6.11 mL of H₂C₂O₄ are needed

Explanation:

We determine the reaction which is a redox one, in acidic medium

C₂O₄⁻² + MnO₄⁻ → CO₂ + Mn²⁺

C₂O₄⁻² → 2CO₂ + 2e⁻ Oxidation

Carbon changes the oxidation state, from +3 to +4

5e⁻ + MnO₄⁻ + 8H⁺ → Mn²⁺ + 4H₂O Reduction

We add 4 water to the product side, in order to balance the oxygen and, we have 8H+ in the reactant side, in order to balance the H

Mn changes the oxidation state from +7 to +2

(C₂O₄⁻² → 2CO₂ + 2e⁻) .5

5C₂O₄⁻² → 10CO₂ + 10e⁻

(5e⁻ + MnO₄⁻ + 8H⁺ → Mn²⁺ + 4H₂O) .2

10e⁻ + 2MnO₄⁻ + 16H⁺ → 2Mn²⁺ + 8H₂O

10e⁻ + 2MnO₄⁻ + 16H⁺ + 5C₂O₄⁻² → 2Mn²⁺ + 8H₂O + 10CO₂ + 10e⁻

The electrons are cancelled, so the balanced reaction is:

2KMnO₄ + 6HCl + 5H₂C₂O₄ → 2MnCl₂ + 8H₂O + 10CO₂ + 2KCl

Concentration of KMnO₄ = 0.11 mol / 0.080mL = 1.375M

Imagine that the reactants are in molar concentration (mol/L)

Ratio in stoichiometry is 2:5

2 moles of KMnO₄ react to 5 moles of oxalic acid

Then, 1.375 moles of KMnO₄ will react to (1.375 moles . 5 )/ 2 = 3.437 M

Concentration of H₂C₂O₄ = 3.437 M (mol/L)

3.437 mol/L = 0.021 moles / Volume (L)

0.021 moles / 3.437 mol/L = Volume (L) → 0.00611 L

0.00611 L . 1000 mL / 1L = 6.11 mL

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3 years ago

A compound is 42.9% C, 2.4% H, 16.7% N, and 38.1% O, by mass. Addition of 6.45 g of this compound to 50.0 mL benzene, lowers the

Romashka [77]

This is an incomplete question, here is a complete question.

A compound is 42.9% C, 2.4% H, 16.7% N and 38.1% O by mass. Addition of 6.45 g of this compound to 50.0 mL benzene, C₆H₆ (d= 0.879 g/mL; Kf= 5.12 degrees Celsius/m), lowers the freezing point from 5.53 to 1.37 degrees Celsius. What is the molecular formula of this compound?

Answer : The molecular of the compound is, $C_6H_4N_2O_4$

Explanation :

First we have to calculate the mass of benzene.

$\text{Mass of benzene}=\text{Density of benzene}\times \text{Volume of benzene}$

$\text{Mass of benzene}=0.879g/mL\times 50.0mL=43.95g$

Now we have to calculate the molar mass of unknown compound.

Given:

Mass of unknown compound (solute) = 6.45 g

Mass of benzene (solvent) = 43.95 g = 0.04395 kg

Formula used :

$\Delta T_f=K_f\times m\\\\\Delta T_f=K_f\times\frac{\text{Mass of unknown compound}}{\text{Molar mass of unknown compound}\times \text{Mass of benzene in Kg}}$

where,

$\Delta T_f$ = change in freezing point = $5.53-1.37=4.16^oC$

$\Delta T_s$ = freezing point of solution

$\Delta T^o$ = freezing point of benzene

Molal-freezing-point-depression constant $(K_f)$ for benzene = $5.12^oC/m$

m = molality

Now put all the given values in this formula, we get

$4.16^oC=(5.12^oC/m)\times \frac{6.45g}{\text{Molar mass of unknown compound}\times 0.04395kg}$

$\text{Molar mass of unknown compound}=180.6g/mol$

If percentage are given then we are taking total mass is 100 grams.

So, the mass of each element is equal to the percentage given.

Mass of C = 42.9 g

Mass of H = 2.4 g

Mass of N = 16.7 g

Mass of O = 38.1 g

Molar mass of C = 12 g/mole

Molar mass of H = 1 g/mole

Molar mass of N = 14 g/mole

Molar mass of O = 16 g/mole

Step 1 : convert given masses into moles.

Moles of C = $\frac{\text{ given mass of C}}{\text{ molar mass of C}}= \frac{42.9g}{12g/mole}=3.575moles$

Moles of H = $\frac{\text{ given mass of H}}{\text{ molar mass of H}}= \frac{2.4g}{1g/mole}=2.4moles$

Moles of N = $\frac{\text{ given mass of N}}{\text{ molar mass of N}}= \frac{16.7g}{14g/mole}=1.193moles$

Moles of O = $\frac{\text{ given mass of O}}{\text{ molar mass of O}}= \frac{38.1g}{16g/mole}=2.381moles$

Step 2 : For the mole ratio, divide each value of moles by the smallest number of moles calculated.

For C = $\frac{3.575}{1.193}=2.99\approx 3$

For H = $\frac{2.4}{1.193}=2.01\approx 2$

For N = $\frac{1.193}{1.193}=1$

For O = $\frac{2.381}{1.193}=1.99\approx 2$

The ratio of C : H : N : O = 3 : 2 : 1 : 2

The mole ratio of the element is represented by subscripts in empirical formula.

The Empirical formula = $C_3H_2N_1O_2$

The empirical formula weight = 3(12) + 2(1) + 1(14) + 2(16) = 84 gram/eq

Now we have to calculate the molecular formula of the compound.

Formula used :

$n=\frac{\text{Molecular formula}}{\text{Empirical formula weight}}$

$n=\frac{180.6}{84}=2$

Molecular formula = $(C_3H_2N_1O_2)_n=(C_3H_2N_1O_2)_2=C_6H_4N_2O_4$

Therefore, the molecular of the compound is, $C_6H_4N_2O_4$

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