Question

Який об'єм водою виділиться при взаємодії 78г калію з водою?​ ПОМОГИТЕ ПОЖАЛУСТА

Answer 1

Если там водню то решение такое, если что-то не понятно, буду рада объяснить

Answer 2

Water Volume Released by the Reaction of 78g of Potassium with Water

When 78g of potassium reacts with water, it produces potassium hydroxide and hydrogen gas. The balanced chemical equation for this reaction is:

2K + 2H2O -> 2KOH + H2

From the balanced equation, we can see that 2 moles of potassium react with 2 moles of water to produce 2 moles of potassium hydroxide and 1 mole of hydrogen gas.

To calculate the volume of hydrogen gas produced, we can use the ideal gas law, which states that 1 mole of any gas at standard temperature and pressure (STP) occupies 22.4 liters.

Calculation: 1. Calculate the moles of potassium: - The molar mass of potassium (K) is approximately 39.10 g/mol. - Number of moles of potassium = Mass of potassium / Molar mass of potassium - Number of moles of potassium = 78g / 39.10 g/mol

2. Calculate the volume of hydrogen gas produced: - According to the balanced equation, 2 moles of potassium produce 1 mole of hydrogen gas. - Therefore, the moles of hydrogen gas produced = 78g / 39.10 g/mol * (1 mole H2 / 2 moles K) - Volume of hydrogen gas produced = Moles of hydrogen gas * 22.4 liters/mol (at STP)

Result: The volume of hydrogen gas produced when 78g of potassium reacts with water is approximately 20.0 liters at standard temperature and pressure (STP).