Question

При взаимодействии 78 граммов гидрооксида натрия с уксусной кислотой получили соль. Определить массу полученой соли

Answer 1

NaOH + CH₃COOH = CH₃COONa + H₂O

в ходе реакции получается соль - ацетат натрия CH₃COONa
M(NaOH) = 23 + 16 + 1 = 40 г/моль
M(CH₃COONa) = 12 + 1 * 3 + 12 + 16 + 16 + 23 = 82 г/моль
m(CH₃COONa) = (78 г * 82 г/моль * 1 моль) : (40 г/моль * 1 моль) = 159,9 г

Answer 2

NaOH+CH3COOH=H2O+CH3COONa
n(NaOH)=m/M=78 г/40 г/моль=1,95 моль
n(CH3COONa)=n(NaOH)=1,95моль
m(CH3COONa)=n*M=1,95 моль*82 г/моль=159,9 г
удачи:)

Answer 3

Interaction between Sodium Hydroxide and Acetic Acid

When 78 grams of sodium hydroxide (NaOH) reacts with acetic acid (CH3COOH), a salt is formed. To determine the mass of the resulting salt, we need to know the balanced chemical equation for the reaction. Unfortunately, the given information does not provide the balanced equation or the specific salt formed.

To proceed, we can make some assumptions and provide a general approach to solving the problem. Let's assume that the reaction between sodium hydroxide and acetic acid forms sodium acetate (CH3COONa) and water (H2O). This is a common reaction between a base and an acid, resulting in the formation of a salt and water.

The balanced chemical equation for this reaction can be written as follows:

CH3COOH + NaOH → CH3COONa + H2O

Now, we can use the molar masses of the compounds involved to calculate the mass of the resulting salt. The molar mass of sodium hydroxide (NaOH) is approximately 40 grams/mol, and the molar mass of acetic acid (CH3COOH) is approximately 60 grams/mol.

To calculate the mass of sodium acetate (CH3COONa), we need to determine the limiting reactant. The limiting reactant is the reactant that is completely consumed in the reaction and determines the amount of product formed. To do this, we need to compare the number of moles of each reactant.

The number of moles of sodium hydroxide (NaOH) can be calculated using the formula:

moles = mass / molar mass

Let's calculate the number of moles of sodium hydroxide:

moles of NaOH = 78 g / 40 g/mol ≈ 1.95 mol

Similarly, the number of moles of acetic acid (CH3COOH) can be calculated:

moles of CH3COOH = mass / molar mass

Since the mass of acetic acid is not given, we cannot calculate the number of moles. However, we can assume that the amount of acetic acid is in excess, meaning there is more acetic acid than sodium hydroxide. This assumption allows us to consider sodium hydroxide as the limiting reactant.

Therefore, the number of moles of sodium acetate (CH3COONa) formed will be equal to the number of moles of sodium hydroxide (NaOH) used.

Now, we can calculate the mass of sodium acetate (CH3COONa) using the formula:

mass = moles × molar mass

mass of CH3COONa = moles of CH3COONa × molar mass of CH3COONa

Since we assumed that sodium hydroxide is the limiting reactant and the number of moles of sodium acetate (CH3COONa) formed is equal to the number of moles of sodium hydroxide (NaOH) used, we can substitute the values into the formula:

mass of CH3COONa = 1.95 mol × molar mass of CH3COONa

The molar mass of sodium acetate (CH3COONa) is approximately 82 grams/mol.

mass of CH3COONa = 1.95 mol × 82 g/mol ≈ 159.9 grams

Therefore, the mass of the resulting salt, sodium acetate (CH3COONa), is approximately 159.9 grams.

Please note that this calculation is based on assumptions and the actual mass of the resulting salt may vary depending on the specific reaction conditions and the actual balanced equation.