Question

Найти массу соли, полученную при взаимодействии уксусной кислоты массой 6 г с избытком гидроксида натрия

Answer 1

CH3COOH+NaOH=CH3COONa+H2O
n(CH3COOH)=6/60=0,1 моль
n(CH2COOH)=n(CH3COONa)
m(CH3COONa)=0,1*82=8,2 г

Answer 2

Calculating the Mass of Salt Produced

To find the mass of salt produced when acetic acid reacts with an excess of sodium hydroxide, we can use the chemical equation for the reaction and the given mass of acetic acid.

The balanced chemical equation for the reaction between acetic acid (CH3COOH) and sodium hydroxide (NaOH) is:

CH3COOH + NaOH → CH3COONa + H2O

From the given information, the mass of acetic acid is 6 g. To calculate the mass of salt produced, we need to determine the amount of acetic acid that reacts and then use stoichiometry to find the mass of the salt produced.

The molar mass of acetic acid (CH3COOH) is approximately 60.05 g/mol, and the molar mass of sodium hydroxide (NaOH) is approximately 40.00 g/mol.

Using this information, we can calculate the mass of salt produced.

The first step is to determine the number of moles of acetic acid present:

Number of moles = Mass / Molar mass

Number of moles of CH3COOH = 6 g / 60.05 g/mol

Number of moles of CH3COOH ≈ 0.1 mol

Now, we need to use the balanced chemical equation to find the number of moles of salt produced. From the balanced equation, we can see that 1 mole of acetic acid reacts with 1 mole of sodium hydroxide to produce 1 mole of sodium acetate (CH3COONa).

Therefore, the number of moles of sodium acetate produced is also approximately 0.1 mol.

Finally, we can calculate the mass of sodium acetate produced:

Mass = Number of moles × Molar mass

Mass of CH3COONa = 0.1 mol × (approximately 82.03 g/mol)

Mass of CH3COONa ≈ 8.2 g

So, the mass of salt produced when 6 g of acetic acid reacts with an excess of sodium hydroxide is approximately 8.2 grams.

Conclusion

The mass of salt produced when acetic acid reacts with an excess of sodium hydroxide is approximately 8.2 grams.