Question

Пожалуйста!!! 1.Соляная кислота взаимодействует с оксидом натрия с образованием хлорида натрия и воды. Определить массу хлорида натрия, полученного при действии соляной кислоты на 3,1 г оксида натрия. Какое количество соляной кислоты вступило в реакцию

Answer 1

Дано:
m (Na2O) = 3.1г

Найти:
m (NaCl) - ?
n (HCl) - ?

Решение:
Na2O + 2HCl = 2NaCl + H2O

Вычислим количество вещества оксида натрия:

n (Na2O) = m/M = 3.1 / 62 = 0.05 моль

По уравнению видно, что количество вещества соляной кислоты и хлорида натрия в 2 раза больше количества вещества оксида натрия:

n (HCl) = 2n (Na2O) = 0.1 моль
n (NaCl) = 2n (Na2O) = 0.1 моль

Вычислим массу хлорида натрия:

m (NaCl) = n * M = 0.1 * 58.5 = 5.85г

Ответ: m (NaCl) = 5.85г; n (HCl) = 0.1 моль

Answer 2

Reaction and Stoichiometry

The reaction between hydrochloric acid (HCl) and sodium oxide (Na2O) forms sodium chloride (NaCl) and water (H2O). We need to determine the mass of sodium chloride produced when 3.1 g of sodium oxide reacts with hydrochloric acid. Additionally, we need to find the amount of hydrochloric acid that reacted.

The balanced chemical equation for the reaction is: 2HCl + Na2O -> 2NaCl + H2O

To find the mass of sodium chloride produced, we need to calculate the molar mass of sodium chloride (NaCl) and use stoichiometry to convert the given mass of sodium oxide (Na2O) to the mass of sodium chloride (NaCl).

Calculation of Mass of Sodium Chloride

The molar mass of sodium chloride (NaCl) is the sum of the atomic masses of sodium (Na) and chlorine (Cl). According to the periodic table, the atomic masses of sodium and chlorine are 22.99 g/mol and 35.45 g/mol, respectively.

Molar mass of NaCl = (1 * atomic mass of Na) + (1 * atomic mass of Cl) = (1 * 22.99 g/mol) + (1 * 35.45 g/mol) = 58.44 g/mol

Using stoichiometry, we can calculate the mass of sodium chloride produced when 3.1 g of sodium oxide reacts.

Moles of Na2O = mass / molar mass = 3.1 g / (2 * atomic mass of Na + atomic mass of O) = 3.1 g / (2 * 22.99 g/mol + 16.00 g/mol) = 3.1 g / 61.98 g/mol ≈ 0.050 mol

According to the balanced equation, 1 mole of Na2O reacts to form 2 moles of NaCl. Therefore, the moles of NaCl produced will be twice the moles of Na2O.

Moles of NaCl = 2 * moles of Na2O = 2 * 0.050 mol = 0.100 mol

Mass of NaCl = moles of NaCl * molar mass of NaCl = 0.100 mol * 58.44 g/mol = 5.844 g

Therefore, the mass of sodium chloride produced when 3.1 g of sodium oxide reacts with hydrochloric acid is approximately 5.844 g.

Calculation of Amount of Hydrochloric Acid

To find the amount of hydrochloric acid that reacted, we need to use stoichiometry and the balanced equation.

According to the balanced equation, 2 moles of hydrochloric acid (HCl) react with 1 mole of sodium oxide (Na2O) to form 2 moles of sodium chloride (NaCl). Therefore, the moles of HCl that reacted will be half the moles of Na2O.

Moles of HCl = 0.5 * moles of Na2O = 0.5 * 0.050 mol = 0.025 mol

To calculate the mass of hydrochloric acid, we need to use the molar mass of HCl, which is the sum of the atomic masses of hydrogen (H) and chlorine (Cl). According to the periodic table, the atomic masses of hydrogen and chlorine are 1.01 g/mol and 35.45 g/mol, respectively.

Molar mass of HCl = (1 * atomic mass of H) + (1 * atomic mass of Cl) = (1 * 1.01 g/mol) + (1 * 35.45 g/mol) = 36.46 g/mol

Mass of HCl = moles of HCl * molar mass of HCl = 0.025 mol * 36.46 g/mol = 0.9115 g

Therefore, the amount of hydrochloric acid that reacted is approximately 0.9115 g.

Please note that the values calculated above are approximate due to the limited precision of the given data.

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