Question

В результате взаимодействия оксида натрия с водой образовалось 8 граммов гидроксида натрия, вычислите массу воды вступившей в реакцию.

Answer 1

2Na+2H2O=2NaOH+H2 вычислим количество вещества гидроксида n(NaOH)=8/(23+1+16)=0,2моль, согласно уравнению реакции столько же воды вступило в реакцию, тогда её масса m(H2O)=0,2x(2+16)=3,6г

Answer 2

Calculating the Mass of Water Reacted with Sodium Oxide

To calculate the mass of water that reacted with sodium oxide to form 8 grams of sodium hydroxide, we can use the balanced chemical equation for the reaction between sodium oxide and water:

2Na2O + 2H2O → 4NaOH

From the balanced equation, we can see that 2 moles of sodium oxide react with 2 moles of water to produce 4 moles of sodium hydroxide.

Using the molar mass of sodium hydroxide (NaOH) and the given mass of sodium hydroxide, we can calculate the moles of sodium hydroxide produced and then determine the moles of water reacted.

Calculation Steps:

1. Calculate the moles of sodium hydroxide produced: - The molar mass of sodium hydroxide (NaOH) is 40.00 g/mol. - Given mass of sodium hydroxide = 8 grams. - Moles of sodium hydroxide = Given mass / Molar mass = 8 g / 40.00 g/mol = 0.2 moles.

2. Determine the moles of water reacted: - From the balanced chemical equation, 2 moles of sodium oxide react with 2 moles of water to produce 4 moles of sodium hydroxide. - Moles of water = Moles of sodium hydroxide = 0.2 moles.

3. Calculate the mass of water reacted: - The molar mass of water (H2O) is 18.02 g/mol. - Mass of water = Moles of water * Molar mass = 0.2 moles * 18.02 g/mol = 3.604 grams.

Answer:

The mass of water that reacted with sodium oxide to form 8 grams of sodium hydroxide is 3.604 grams.