Na2O+H2O=? уравнение реакции Какие продукты образуются в результате взаимодействия Na2O + H2O = ? Надо записать молекулярное уравнение реакции, а также решить задачу. Вот условие: какая масса гидроксида натрия образуется при растворении 24 г оксида натрия в 75 г воды?
So, when 24 g of sodium oxide (Na2O) reacts with water (H2O), it forms 2 moles of sodium hydroxide (NaOH).
First, we need to calculate the molar mass of Na2O and NaOH: Na2O: Na (Sodium) = 22.99 g/mol * 2 + O (Oxygen) = 16.00 g/mol = 61.98 g/mol NaOH: Na (Sodium) = 22.99 g/mol + O (Oxygen) = 16.00 g/mol + H (Hydrogen) = 1.01 g/mol = 39.99 g/mol
Next, we calculate the moles of Na2O: 24 g / 61.98 g/mol = 0.387 moles
Since the reaction produces 2 moles of NaOH for every 1 mole of Na2O, the moles of NaOH produced will be: 0.387 moles * 2 = 0.774 moles
Now, we calculate the mass of NaOH produced: 0.774 moles * 39.99 g/mol = 30.89 g
Therefore, the mass of sodium hydroxide formed when 24 g of sodium oxide is dissolved in 75 g of water is 30.89 grams.
Molecular equation for the reaction:
Na2O + H2O -> 2NaOH
So, when 24 g of sodium oxide (Na2O) reacts with water (H2O), it forms 2 moles of sodium hydroxide (NaOH).
First, we need to calculate the molar mass of Na2O and NaOH:
Na2O: Na (Sodium) = 22.99 g/mol * 2 + O (Oxygen) = 16.00 g/mol = 61.98 g/mol
NaOH: Na (Sodium) = 22.99 g/mol + O (Oxygen) = 16.00 g/mol + H (Hydrogen) = 1.01 g/mol = 39.99 g/mol
Next, we calculate the moles of Na2O:
24 g / 61.98 g/mol = 0.387 moles
Since the reaction produces 2 moles of NaOH for every 1 mole of Na2O, the moles of NaOH produced will be:
0.387 moles * 2 = 0.774 moles
Now, we calculate the mass of NaOH produced:
0.774 moles * 39.99 g/mol = 30.89 g
Therefore, the mass of sodium hydroxide formed when 24 g of sodium oxide is dissolved in 75 g of water is 30.89 grams.