Question

ДАЮ 100 БАЛЛОВ!! СРОЧНО!! Який об'єм водню виділиться і яка маса солі утвориться при взаємодії 450 г розчину етанової кислоти з масовою часткою кислоти 30% з магнієм?

Answer 1

Ответ:

Объяснение:

Дано:

m(р-нCH3COOH) = 450 г

w(CH3COOH) = 30% або 0,3

Найти: V(H2) - ?  m(Mg(CH3COO)2) - ?

Знайдемо масу етанової кислоти у розчині:

m(CH3COOH) = m(р-нCH3COOH) * w(CH3COOH) = 450 * 0,3 = 135 г

Складемо РХР:

               135 г                   х г                  у л

Mg + 2 CH3COOH → Mg(CH3COO)2 + H2

         n = 2 моль         n = 1 моль           n= 1 моль

Знайдемо кількість речовини CH3COOH

М(CH3COOH) = 12 * 2 + 1 * 4 + 16 * 2 = 60 г/моль

n(CH3COOH) = m(CH3COOH)/М(CH3COOH) = 135/60 = 2,25 моль

З РХР бачимо, що n(CH3COOH) : n(Mg(CH3COO)2) = 2 : 1 звідси маємо

n(Mg(CH3COO)2) = 1/2n(Mg(CH3COO)2) = 1/2 * 2,25 = 1,125 моль

М(Mg(CH3COO)2) = 24 + 12 * 2 + 1 * 6 + 16 * 2 = 86 г/моль

m(Mg(CH3COO)2) = n(Mg(CH3COO)2) * М(Mg(CH3COO)2) =

= 1,125 * 86 = 96,75 г

З РХР бачимо, що n(CH3COOH) : n(Н2) = 2 : 1 звідси маємо

n(Н2) = 1/2n(CH3COOH) = 1/2 * 2,25 = 1,125 моль

V(H2) = n(Н2) * Vm = 1,125 * 22,4 = 25,2 л

Answer 2

To calculate the volume of hydrogen gas released and the mass of salt formed during the reaction between 450 g of a solution of ethanoic acid with a mass fraction of acid of 30% and magnesium, we need to use the balanced chemical equation for the reaction.

The balanced chemical equation for the reaction between ethanoic acid (CH3COOH) and magnesium (Mg) is:

2CH3COOH + Mg -> (CH3COO)2Mg + H2

From the balanced equation, we can see that 2 moles of ethanoic acid react with 1 mole of magnesium to produce 1 mole of magnesium ethanoate and 1 mole of hydrogen gas.

Now, let's calculate the moles of ethanoic acid and magnesium in the given quantities:

1. Moles of ethanoic acid: - Mass of ethanoic acid = 450 g - Mass fraction of acid = 30% - Mass of acid = (30/100) * 450 g = 135 g - Molar mass of ethanoic acid (CH3COOH) = 60.052 g/mol - Moles of ethanoic acid = 135 g / 60.052 g/mol

2. Moles of magnesium: - Mass of magnesium = Mass of solution - Mass of acid - Mass of solution = 450 g - Mass of magnesium = 450 g - 135 g - Molar mass of magnesium (Mg) = 24.305 g/mol - Moles of magnesium = Mass of magnesium / Molar mass of magnesium

Now that we have the moles of ethanoic acid and magnesium, we can determine the limiting reactant and calculate the volume of hydrogen gas and the mass of salt formed.

The limiting reactant is the reactant that is completely consumed and determines the amount of product formed. It can be determined by comparing the moles of the reactants using the stoichiometric coefficients from the balanced equation.

In this case, the stoichiometric coefficient of ethanoic acid is 2, and the stoichiometric coefficient of magnesium is 1. Therefore, the moles of ethanoic acid should be divided by 2 to compare them correctly.

Let's calculate the moles of hydrogen gas and magnesium ethanoate formed based on the limiting reactant:

1. Moles of hydrogen gas: - Moles of ethanoic acid (divided by 2) = Moles of hydrogen gas - Moles of hydrogen gas = Moles of ethanoic acid / 2

2. Moles of magnesium ethanoate: - Moles of magnesium = Moles of magnesium ethanoate

Now, we can use the ideal gas law to calculate the volume of hydrogen gas released. Assuming the reaction takes place at standard temperature and pressure (STP), the molar volume of an ideal gas is 22.4 L/mol.

3. Volume of hydrogen gas: - Volume of hydrogen gas = Moles of hydrogen gas * Molar volume at STP

Finally, to calculate the mass of magnesium ethanoate formed, we need to multiply the moles of magnesium ethanoate by its molar mass.

4. Mass of magnesium ethanoate: - Molar mass of magnesium ethanoate ((CH3COO)2Mg) = 142.394 g/mol - Mass of magnesium ethanoate = Moles of magnesium ethanoate * Molar mass of magnesium ethanoate

Please note that I will now calculate the values based on the given information.