Question

Какой объем кислорода (н.у.) потребуется для получения 76,5 грамм оксида бария ?

Answer 1

Ba2 + O2 = BaO
V=Vm * n
V= 22.4 * 1 моль = 22.4 л
M(BaO) = 16+137=153 г/моль
m(BaO ) = 153г
x:22.4=76.5:153
x= 76.5*22.4:153=11.2 л
Ответ 11.2 л

Answer 2

Calculating Oxygen Volume for Barium Oxide

To calculate the volume of oxygen gas (at standard temperature and pressure) required to produce 76.5 grams of barium oxide, we can use stoichiometry and the molar mass of barium oxide.

1. Molar Mass of Barium Oxide: - The molar mass of barium oxide (BaO) is 153.33 g/mol.

2. Stoichiometry: - The balanced chemical equation for the formation of barium oxide from its elements is: `2 Ba + O2 → 2 BaO`

3. Calculating Oxygen Volume: - First, we need to find the number of moles of barium oxide using its molar mass: `moles = mass / molar mass` `moles = 76.5 g / 153.33 g/mol = 0.5 moles` - According to the balanced chemical equation, 1 mole of barium oxide requires 1 mole of oxygen gas. - Therefore, 0.5 moles of barium oxide will require 0.5 moles of oxygen gas. - Using the ideal gas law, we can calculate the volume of oxygen gas at STP: `PV = nRT`, where P = 1 atm, V = volume, n = moles, R = 0.0821 L·atm/(K·mol), and T = 273 K. `V = nRT / P` `V = (0.5 moles) * (0.0821 L·atm/(K·mol)) * (273 K) / (1 atm)` `V ≈ 11.2 liters`

Therefore, approximately 11.2 liters of oxygen gas at STP will be required to produce 76.5 grams of barium oxide.

I hope this helps! If you have further questions, feel free to ask!