Question

Определите объем кислорода, необходимый для полного сжигания 3-х моль кислорода (н.у.).

Answer 1

N(O2)=3 моль
V(O2)=3*22.4=67.2 л

Answer 2

Calculation of the Volume of Oxygen Required for the Complete Combustion of 3 Moles of Oxygen (NTP)

To determine the volume of oxygen required for the complete combustion of 3 moles of oxygen at standard temperature and pressure (NTP), we can use the ideal gas law equation:

PV = nRT

Where: - P is the pressure (at NTP, it is approximately 1 atmosphere) - V is the volume of the gas - n is the number of moles of the gas - R is the ideal gas constant (approximately 0.0821 L·atm/(mol·K)) - T is the temperature (at NTP, it is approximately 273.15 K)

In this case, we are looking for the volume of oxygen (V) when the number of moles (n) is 3.

Let's substitute the values into the equation:

V = (nRT) / P

V = (3 moles) x (0.0821 L·atm/(mol·K)) x (273.15 K) / (1 atm)

Calculating this expression gives us the volume of oxygen required for the complete combustion of 3 moles of oxygen at NTP.

Please note that the question seems to be asking for the volume of oxygen required to combust 3 moles of oxygen, which is a bit confusing. It is not possible to combust oxygen with oxygen itself. If you meant to ask about the combustion of another substance using oxygen, please provide more details.

Let me know if you need any further assistance!