Question

оксид натрия массой 115г растворился в воде определите количество вещества полученного гидроксида натрия

Answer 1

Во вложении:

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Answer 2

To determine the amount of sodium hydroxide (NaOH) produced when 115 grams of sodium oxide (Na2O) dissolves in water, we first need to write the balanced chemical equation for the reaction.

The reaction between sodium oxide and water (H2O) forms sodium hydroxide:

Na2O + H2O → 2NaOH

From the balanced equation, we can see that 1 mole of sodium oxide (Na2O) reacts to produce 2 moles of sodium hydroxide (NaOH).

To calculate the amount of sodium hydroxide formed, we need to follow these steps:

Step 1: Calculate the molar mass of Na2O: Na (sodium) has a molar mass of 22.99 g/mol. O (oxygen) has a molar mass of 16.00 g/mol. Molar mass of Na2O = 2 * (Na's molar mass) + 1 * (O's molar mass) Molar mass of Na2O = 2 * 22.99 + 1 * 16.00 = 61.98 g/mol

Step 2: Calculate the number of moles of Na2O: Number of moles = Mass / Molar mass Number of moles of Na2O = 115 g / 61.98 g/mol ≈ 1.854 moles

Step 3: Calculate the number of moles of NaOH formed: From the balanced equation, 1 mole of Na2O produces 2 moles of NaOH. Number of moles of NaOH = 2 * Number of moles of Na2O Number of moles of NaOH = 2 * 1.854 moles ≈ 3.708 moles

So, approximately 3.708 moles of sodium hydroxide (NaOH) are produced when 115 grams of sodium oxide (Na2O) dissolves in water.