Question

Який об’єм водню (н. у.) виділиться під час взаємодії 3 г натрію з розбавленою сульфатною кислотою? ДАЮ 100 БАЛЛОВ

Answer 1

Ответ:

Об'єм водню - 2,92 літрів

Answer 2

Reaction of Sodium with Dilute Sulfuric Acid

The reaction of 3 grams of sodium with dilute sulfuric acid can be calculated to determine the volume of hydrogen gas produced.

Calculation: The balanced chemical equation for the reaction between sodium and sulfuric acid is: 2Na + H2SO4 -> Na2SO4 + H2

From the balanced equation, it is clear that 2 moles of sodium react with 1 mole of sulfuric acid to produce 1 mole of hydrogen gas.

The molar mass of sodium is 22.99 g/mol, and the molar mass of sulfuric acid (H2SO4) is 98.08 g/mol.

Using the molar masses, we can calculate the moles of sodium and sulfuric acid involved in the reaction: - Moles of sodium = 3g / 22.99 g/mol - Moles of sulfuric acid = (3g / 22.99 g/mol) / 2

Now, we can use the mole ratio from the balanced equation to find the moles of hydrogen gas produced and then convert it to volume using the ideal gas law.

Let's calculate the moles of hydrogen gas and then the volume of hydrogen gas produced.

Search Result: According to the search result from You.com, the volume of hydrogen gas produced can be calculated using the ideal gas law, which states that PV = nRT, where P is the pressure, V is the volume, n is the number of moles, R is the gas constant, and T is the temperature.

The molar volume of an ideal gas at standard temperature and pressure (STP) is 22.4 liters per mole.

Using this information, we can calculate the volume of hydrogen gas produced during the reaction of 3 grams of sodium with dilute sulfuric acid.