Question

Рассчитайте массу 10 процентного раствора гидроксида натрия необходимого для полной нейтрализации 50 г смеси муравьиной и уксусной кислот массовая доля уксусной кислоты в которой составляет 60%​

Answer 1

Ответ:

m(CH3COOH)=50*0.6=30 грамм

n(CH3COOH)=30/60=0,5 моль

m(HCOOH)=50-30=20 грамм

n(HCOOH)=20/46=0,43 моль

CH3COOH + NaOH = СH3COONa + H2O

n(CH3COOH)=n(NaOH)=0.5 моль

HCOOH + NaOH = HCOONa + H2O  

n(HCOOH)=n(NaOH)=0,43 моль

Итого n(NAOH)=0.5+0.43=0.93 моль

m(NаOH)=0,93*40=37,2 грамм

m(NAOH 10%)=37.2/0.1=372 грамм

Answer 2

Ответ:

ответ: 612.5г

Объяснение:

Масса расствора 612.5г

Answer 3

Calculation of the Mass of 10% Sodium Hydroxide Solution Required for Neutralizing a Mixture of Formic and Acetic Acid

To calculate the mass of a 10% sodium hydroxide (NaOH) solution required for the complete neutralization of a mixture of formic and acetic acid, we need to consider the following information:

- The mass of the mixture of formic and acetic acid is 50 g. - The mass fraction of acetic acid in the mixture is 60%.

To solve this problem, we can follow these steps:

1. Calculate the mass of acetic acid in the mixture: - Mass of acetic acid = mass of the mixture × mass fraction of acetic acid - Mass of acetic acid = 50 g × 0.6 = 30 g

2. Calculate the molar mass of acetic acid (CH3COOH): - The molar mass of carbon (C) = 12.01 g/mol - The molar mass of hydrogen (H) = 1.01 g/mol - The molar mass of oxygen (O) = 16.00 g/mol - The molar mass of acetic acid = (2 × molar mass of carbon) + (4 × molar mass of hydrogen) + molar mass of oxygen - The molar mass of acetic acid = (2 × 12.01) + (4 × 1.01) + 16.00 = 60.05 g/mol

3. Calculate the number of moles of acetic acid: - Number of moles of acetic acid = mass of acetic acid / molar mass of acetic acid - Number of moles of acetic acid = 30 g / 60.05 g/mol = 0.4999 mol (approximately 0.5 mol)

4. Since sodium hydroxide (NaOH) has a 1:1 stoichiometric ratio with acetic acid (CH3COOH) in the neutralization reaction, the number of moles of sodium hydroxide required will also be 0.5 mol.

5. Calculate the mass of sodium hydroxide required: - Mass of sodium hydroxide = number of moles of sodium hydroxide × molar mass of sodium hydroxide - The molar mass of sodium (Na) = 22.99 g/mol - The molar mass of oxygen (O) = 16.00 g/mol - The molar mass of hydrogen (H) = 1.01 g/mol - The molar mass of sodium hydroxide = molar mass of sodium + molar mass of oxygen + molar mass of hydrogen - The molar mass of sodium hydroxide = 22.99 + 16.00 + 1.01 = 40.00 g/mol - Mass of sodium hydroxide = 0.5 mol × 40.00 g/mol = 20 g

Therefore, the mass of a 10% sodium hydroxide solution required for the complete neutralization of the mixture of formic and acetic acid is 20 grams.

Please note that the calculation assumes ideal conditions and a complete reaction.