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3 years ago

What is the molarity of the potassium hydroxide if 27.20 mL of KOH is required to neutralize 0.604 g of oxalic acid, H2C2O4?

Chemistry

1 answer:

kakasveta [241]3 years ago

6 0

Answer:

Concentration of KOH = 1.154 M

Explanation:

$H_2C_2O_4(aq) + 2KOH(aq) \rightarrow K_2C_2O_4(aq) + 2H_2O(l)$

In the above reaction, 1 mole of oxalic acid reacts with 2 moles of KOH.

Mass of oxalic acid = 0.604 g

$Mole = \frac{Mass\; in\;g}{Molecular\;mass}$

Molecular mass of oxalic acid = 90.03 g/mol

$Mole = \frac{0.604}{90.03}=0.0067\;mol$

1 mol of oxalic acid reacts with 2 moles of KOH

0.0067 mol of oxalic acid reacts with $0.0067\times 2 = 0.0134 mol\; of\;KOH$

Volume of the solution = 27.02 mL = 0.0272 L

$Molarity=\frac{Mole}{Volume\;in\;L}$

No. of mole of KOH = 0.0134 mol

$Molarity=\frac{0.0134}{0.0272}=1.154\;M$

Concentration of KOH = 1.154 M

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Question 3. A batch chemical reactor achieves a reduction in

kotykmax [81]

Answer:

Rate constant for zero-order kinetics: 1, 58 [mg/L.s]

Rate constant for first-order kinetics: 0,05 [1/s]

Explanation:

The reaction order is the relationship between the concentration of species and the rate of the reaction. The rate law is as follows:

$r = k [A]^{x} [B]^{y}$

where:

[A] is the concentration of species A,
x is the order with respect to species A.
[B] is the concentration of species B,
y is the order with respect to species B

k is the rate constant

The concentration time equation gives the concentration of reactants and products as a function of time. To obtain this equation we have to integrate de velocity law:

$v(t) = -\frac{d[A]}{dt} = k [A]^{n}$

For the kinetics of zero-order, the rate is apparently independent of the reactant concentration.

Rate Law: rate = k

Concentration-time Equation: [A]=[A]o - kt

where

k: rate constant [M/s]

[A]: concentration in the time t [M]
[A]o: initial concentration [M]
t: elapsed reaction time [s]

For first-order kinetics, we have:

Rate Law: rate= k[A]

Concentration -Time Equation: ln[A]=ln[A]o - kt

where:

K: rate constant [1/s]
ln[A]: natural logarithm of the concentration in the time t [M]
ln[A]o: natural logarithm of the initial concentration [M]
t: elapsed reaction time [s]

To solve the problem, wee have the following data:

[A]o = 100 mg/L

[A] = 5 mg/L

t = 1 hour = 60 s

As we don't know the molar mass of the compound A, we can't convert the used concentration unit (mg/L) to molar concentration (M). So we'll solve the problem using mg/L as the concentration unit.

Zero-order kinetics

we use: [A]=[A]o - Kt

we replace the data: 5 = 100 - K (60)

we clear K: K = [100 - 5 ] (mg/L) /60 (s) = 1, 583 [mg/L.s]

First-order kinetics

we use: ln[A]=ln[A]o - Kt

we replace the data: ln(5) = ln(100) - K (60)

we clear K: K = [ln(100) - ln(5)] /60 (s) = 0,05 [1/s]

4 0

3 years ago

How many moles of hydrogen gas will be produced when 12 g of Mg will react completely with excess of an acid according to the fo

lina2011 [118]

Answer:

0.49 mol

Explanation:

Step 1: Write the balanced equation

Mg + 2 HCI ⇒ MgCl₂ + H₂

Step 2: Calculate the moles corresponding to 12 g of Mg

The molar mass of Mg is 24.31 g/mol.

$12g \times \frac{1mol}{24.31g} = 0.49mol$

Step 3: Calculate the moles of H₂ produced by 0.49 moles of Mg

The molar ratio of Mg to H₂ is 1:1. The moles of H₂ produced are 1/1 × 0.49 mol = 0.49 mol.

7 0

3 years ago

2. Water is an example of a(n)

Alexus [3.1K]

it would be A ,inorganic Compound

6 0

2 years ago

This question has multiple parts. Work all the parts to get the most points. An aqueous antifreeze solution is 39.0% ethylene gl

nlexa [21]

Answer:

Molality = 10.300 m

Molarity = 6.5970 M

mole fraction = 0.156549

Explanation:

39.0 % = ethylene glycol

61.0 % = water

imagine the total mass = 100g

39.0% ethylene glycol = 39g

61.0 % water = 61g

1) Molality = number of moles / mass of solvent (kg)

Molar mass of ethylene glycol = 62.07g/mole

mole = 39g / 62.07g/mole = 0.6283 moles

Molality = moles / mass of solvent = 0.6283 moles / 0.061kg = 10.300 m

2) Molarity = number of moles / volume of solution

Since we know the density of the solution = 1.05g /ml

⇒ volume = 100g / 1.05g /mL = 95.24 mL = 0.09524 L

Molarity = 0.6283 moles / 0.09524 L = 6.5970 M

3) Mole fraction

moles water = 61g / 18.02g/mole = 3.38513 moles

Total number of moles = moles of ethylene glycol + moles of water = 0.6283 + 3.38513 = 4.0134276 moles

Mole fraction = 0.6283/ 4.0134276 = 0.156549

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3 years ago

The solubility of oxygen gas in water at 40 ∘c is 1.0 mmol/l of solution. What is this concentration in units of mole fraction?

juin [17]

The formula for mole fraction is:

$mole fraction of solute = \frac{number of moles of solute}{total number of moles of solution}$ -(1)

The solubility of oxygen gas = 1.0 mmol/L (given)

1.0 mmol/L means 1.0 mmol are present in 1 L.

Converting mmol to mol:

$1.00 mmol\times \frac{1 mol}{1000 mmol} = 0.001 mol$

So, moles of oxygen = 0.001 mol

For moles of water:

1 L of water = 1000 mL of water

Since, the density of water is 1.0 g/mL.

$Density = \frac{mass}{volume}$

$Mass = 1.0 g/ml\times 1000 mL = 1000 g$

So, the mass of water is 1000 g.

Molar mass of water = 18 g/mol.

Number of moles of water = $\frac{1000 g}{18 g/mol} = 55.55 mol$

Substituting the values in formula (1):

$mole fraction = \frac{0.001}{55.55+0.001}$

$mole fraction = 1.8\times 10^{-5}$

Hence, the mole fraction is $1.8\times 10^{-5}$ .

7 0

3 years ago