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2 years ago

Which reaction occurs when equivalent quantities of H" (or H30) and OH" are mixed?

1 answer:

4 0

Answer: A reaction which occurs when equivalent quantities of H" (or $H_{3}O^{+}$ ) and OH" are mixed is a neutralization reaction.

Explanation:

When acid and base chemically combine together to form sat and water then it is know as neutralization reaction.

An acidic substance gives hydrogen ions $(H^{+})$ on dissociation and a basic substance gives hydroxide ions $(OH^{-})$ on dissociation.

For example, $H^{+} + OH^{-} \rightarrow H_{2}O$ is a neutralization reaction where an acid and base are mixed together to form water.

No salt is formed as because here the reactants are not present as compounds.

In a hydrolysis reaction, water molecule(s) is added to the reactant molecules.

For example, $CH_{3}COOCH_{3} + H_2O \rightarrow CH_{3}COOH + CH_{3}OH$

Oxidation is the process of addition of oxygen to a molecule and removal of hydrogen.

Reduction is the process of addition of hydrogen in a compound and the removal of oxygen atom.

Therefore, we can conclude that reaction which occurs when equivalent quantities of H" (or $H_{3}O^{+}$ ) and OH" are mixed is a neutralization reaction.

You might be interested in

Suppose of potassium sulfate is dissolved in of a aqueous solution of sodium chromate. Calculate the final molarity of potassium

dimulka [17.4K]

Answer:

This question is incomplete, here's the complete question:

"Suppose 0.0842g of potassium sulfate is dissolved in 50.mL of a 52.0mM aqueous solution of sodium chromate. Calculate the final molarity of potassium cation in the solution. You can assume the volume of the solution doesn't change when the potassium sulfate is dissolved in it. Round your answer to 2 significant digits."

Explanation:

Reaction :-

K2SO4 + Na2CrO4 ------> K2CrO4 + Na2SO4

Mass of K2SO4 = 0.0842 g, Molar mass of K2SO4 = 174.26 g/mol

Number of moles of K2SO4 = 0.0842 g / 174.26 g/mol = 0.000483 mol

Concentration of Na2CrO4 = 52.0 mM = 52.0 * 10^-3 M = 0.052 mol/L

Volume of Na2CrO4 solution = 50.0 ml = 50 L / 1000 = 0.05 L

Number of moles of Na2CrO4 = 0.05 L * 0.052 mol/L = 0.0026 mol

Since number of moles of K2SO4 is smaller than number of moles Na2CrO4, so 0.000483 mol of K2SO4 will react with 0.000483 mol of Na2CrO4 will produce 0.000483 mol of K2CrO4.

0.000483 mol of K2CrO4 will dissociate into 2* 0.000483 mol of K^+

Final concentration of potassium cation

= (2*0.000483 mol) / 0.05 L = 0.02 mol/L = 0.02 M

8 0

3 years ago

S8 + 24 F2 ⟶ 8 SF6

Arturiano [62]

Answer:

Theoretical Yield of SF₆ = 2.01 moles

Explanation: If you understand and can apply the methodology below, you will find it applies to ALL chemical reaction stoichiometry problems based on the balanced standard equation; i.e., balanced to smallest whole number coefficients.

Solution 1:

Rule => Convert given mass values to moles, solve problem using coefficient ratios. Finish by converting moles to the objective dimensions.

Given S₈ + 24F₂ => 8SF₆

425g 229g ?

= 425g/256g/mol. = 226g/38g/mol.

= 1.66 moles S₈ = 6.03 moles F₂ <= Limiting Reactant

Determining Limiting Reactant => Divide moles each reactant by their respective coefficient; the smaller value will always be the limiting reactant.

S₈ = 1.66/1 = 1.66

F₂ = 6.03/24 = 0.25 => F₂ is the limiting reactant

Determining Theoretical Yield:

Note: When working problem do not use the division ratio results for determining limiting reactant. Use the moles F₂ calculated from 229 grams F₂ => 6.03 moles F₂. The division procedure to define the smaller value and limiting reactant is just a quick way to find which reactant controls the extent of reaction.

Given S₈ + 24F₂ => 8SF₆

425g 229g ?

= 425g/256g/mol. = 226g/38g/mol.

= 1.66 moles S₈ = 6.03 moles F₂ <= Limiting Reactant

Max #moles SF₆ produced from 6.03 moles F₂ and an excess S₈

Since coefficient values represent moles, the reaction ratio for the above reaction is 24 moles F₂ to 8 moles SF₆. Such implies that the moles of SF₆ (theoretical) calculated from 6.03 moles of F₂ must be a number less than the 6.03 moles F₂ given. This can be calculated by using a ratio of equation coefficients between 24F₂ and 8SF₆ to make the outcome smaller than 6.03. That is,

moles SF₆ = 8/24 x 6.03 moles = 2.01 moles SF₆ (=> theoretical yield)

S₈ + 24F₂ => 8SF₆

moles SF₆ = 8/24(6.03) moles = 2.01 moles

You would NOT want to use 24/8(6.03) = 18.1 moles which is a value >> 6.03.

This analysis works for all reaction stoichiometry problems.

Convert to moles => divide by coefficients for LR => solve by mole mole ratios from balanced reaction and moles of given.

____________________

Here's another example just for grins ...

C₂H₆O + 3O₂ => 2CO₂ + 3H₂O

Given: 253g 307g ? ?

a. Determine Limiting Reactant

b. Determine mass in grams of CO₂ & H₂O produced

Limiting Reactant

moles C₂H₆O = 253g/46g/mol = 5.5 moles => 5.5/1 = 5.5

moles O₂ = 307g/32g/mol = 9.6 moles => 9.6/24 = 0.4 ∴ O₂ is L.R.

But the problem is worked using the mole values; NOT the number results used to ID the limiting reactant.

C₂H₆O + 3O₂ => 2CO₂ + 3H₂O

------------ 9.6 mole (L.R.) ? ?

mole yield CO₂ = 2/3(9.6)mole = 6.4 mole (CO₂ coefficient < O₂ coefficient)

mole yield H₂O = 9.6mole = 9.6mole (coefficients O₂ & CO₂ are same.)

mole used C₂H₆O = 1/3(9.6)mole = 3.2 mole (coefficient C₂H₆O < coefficient O₂)

For grams => moles x formula weight (g/mole)

7 0

3 years ago

Which statements describes Newton's law of universal gravitation ?

IrinaK [193]

Your answer is D. You're Welcome

8 0

3 years ago

Read 2 more answers

A student was given a stock solution with a concentration of 3.61 g/mL and asked to perform

vovangra [49]

Answer:

The concentration of the dilute sample will be 0.361 g/ml

Explanation:

If a solution is diluted into 1:10 ratio then the amount of solute of that solution will be decreased by 10 times.

The initial concentration of the stock solution was 3.61g/ml but when the solution is diluted in 1:10 ratio the solute concentration is also decreased by 10 times.SO at present the solute concentration becomes 3.61/10=0.361 g/ml.

8 0

3 years ago

How many grams of MgCl2 would be needed to make 5.3 liters of a 0.25 molar solution? Round your answer to the nearest 0.01,

GuDViN [60]

Answer: 126.1 grams of are needed to make 5.3 L of 0.25 M solution

Explanation:

Molarity of a solution is defined as the number of moles of solute dissolved per liter of the solution.

To calculate the mass of solute for given molarity, we use the equation:

$\text{Molarity of the solution}=\frac{\text{given mass of solute}}{\text {Molar mass of solute}\times \text{Volume of solution in L}}$