Question

Який обєм водню виділяється при взаємодії етилегліколю масою 4г з металічним натрієм масою 2,3г

Answer 1

M(C2H6O2) = 60 г/моль
n(C2H6O2) = 4 г/60 г/моль = 0,06667 моль
M(Na) = 23 г/моль
n(Na) = 2.3 г/23 г/моль = 0,1 моль

CH2OH-CH2OH + 2Na => CH2ONa+CH2ONa + H2
Із стехіометричних коефіцієнтів рівняння реакції слідує, що етиленгликоль взят в надлишку, тому
n(H2) = 1/2n(Na) = 0,1 моль/2=0,05 моль
Vm = 22,4 л/моль
V(H2) = 0.05 моль*22,4 л/моль = 1,12 л
Відповідь: 1,12 л

Answer 2

Calculating the Volume of Hydrogen Gas Produced

To calculate the volume of hydrogen gas produced when ethylene glycol with a mass of 4g reacts with metallic sodium with a mass of 2.3g, we can use the following chemical equation for the reaction:

C2H6O2 + 2Na → 2NaOH + H2

From the balanced chemical equation, we can see that 1 mole of ethylene glycol reacts with 2 moles of sodium to produce 1 mole of hydrogen gas.

Step 1: Calculate the Moles of Ethylene Glycol and Sodium

The molar mass of ethylene glycol (C2H6O2) is approximately 62.07 g/mol, and the molar mass of sodium (Na) is approximately 22.99 g/mol.

The number of moles of ethylene glycol (C2H6O2) can be calculated using the formula:

Number of moles = Mass / Molar mass

For ethylene glycol: Number of moles = 4g / 62.07 g/mol

For sodium: Number of moles = 2.3g / 22.99 g/mol

Step 2: Determine the Limiting Reagent

The reactant that produces the least amount of product is the limiting reagent. In this case, we compare the moles of ethylene glycol and sodium to determine the limiting reagent.

Step 3: Calculate the Volume of Hydrogen Gas Produced

Once we have identified the limiting reagent, we can use the stoichiometry of the balanced chemical equation to calculate the volume of hydrogen gas produced.

I will now calculate the moles of ethylene glycol and sodium and determine the limiting reagent.