Question

Яка кількість речовини натрій гідроксиду утворюється під час реакції 120 г натрій оксиду з водою?

Answer 1

1) Уравнение реакции:

Na₂O + H₂O → 2NaOH

2) Находим молярную массу натрий оксида:

М(Na₂O) = 2*23 + 16 = 62 г/моль

3) Анализируем уравнение реакции: из 1 моль Na₂O ( а это 62 г ) образуется 1 моль NaОН. Составляем пропорцию:

62 г Na₂O ----------------- 1 моль NaОН

120 г Na₂O ----------------- Х моль NaОН

Х = 120*1/62 = 1,9 моль

Answer 2

Calculation of Sodium Hydroxide Formation

To calculate the amount of sodium hydroxide formed during the reaction between 120 g of sodium oxide and water, we need to consider the balanced chemical equation for the reaction. The balanced equation for the reaction between sodium oxide (Na2O) and water (H2O) to form sodium hydroxide (NaOH) is as follows:

2 Na2O + 2 H2O -> 4 NaOH

According to the balanced equation, 2 moles of sodium oxide react with 2 moles of water to produce 4 moles of sodium hydroxide. From the equation, we can see that the molar ratio between sodium oxide and sodium hydroxide is 2:4, which simplifies to 1:2.

To find the amount of sodium hydroxide formed, we can use the molar mass of sodium oxide (61.98 g/mol) and the molar mass of sodium hydroxide (39.997 g/mol) to convert the given mass of sodium oxide to moles, and then use the molar ratio to calculate the moles of sodium hydroxide formed.

1. Convert the mass of sodium oxide to moles: - Molar mass of sodium oxide (Na2O) = 61.98 g/mol - Moles of sodium oxide = mass / molar mass = 120 g / 61.98 g/mol

2. Calculate the moles of sodium hydroxide formed: - Moles of sodium hydroxide = Moles of sodium oxide * (2 moles of NaOH / 2 moles of Na2O)

3. Convert the moles of sodium hydroxide to grams: - Mass of sodium hydroxide = Moles of sodium hydroxide * molar mass of sodium hydroxide

Let's calculate the amount of sodium hydroxide formed using the given information:

1. Convert the mass of sodium oxide to moles: - Molar mass of sodium oxide (Na2O) = 61.98 g/mol - Moles of sodium oxide = 120 g / 61.98 g/mol = 1.935 mol (rounded to three decimal places)

2. Calculate the moles of sodium hydroxide formed: - Moles of sodium hydroxide = 1.935 mol * (2 mol NaOH / 2 mol Na2O) = 1.935 mol

3. Convert the moles of sodium hydroxide to grams: - Molar mass of sodium hydroxide (NaOH) = 39.997 g/mol - Mass of sodium hydroxide = 1.935 mol * 39.997 g/mol = 77.88 g (rounded to two decimal places)

Therefore, during the reaction between 120 g of sodium oxide and water, approximately 77.88 grams of sodium hydroxide is formed.

Please note that the above calculation assumes the reaction goes to completion and there are no other limiting factors or side reactions involved.