Question

При полном сгорании метана СН4 в кислороде выделилось 3,6 г воды. Вычислите (при н.у.) объём вступившего в реакцию кислорода. (Ответ должен получиться 4,48л) Нужно объяснить как это делать.

Answer 1

Ответ:

для начала пишем уравнение реакции

CH4 + 2 O2 = CO2 + 2 H2O

затем находим количество вещества (ню) у воды

n=m/M

ищем молярную массу воды

М(H2O)= 1×2 +16=18 г/моль

затем находим ню, поделив известную нам массу на молярную

n=3,6/18=0.2 моль

затем смотрим на коэффициенты перед водой и кислородом(объем которого нужно найти). они равны, потому

n(H2O) = n (O2) = 0,2 моль

далее по формуле V= n× Vm ( где Vm=22,4 л/моль) ищем объем кислорода

V= 0,2×22,4=4,48 л

Answer 2

Calculation of Oxygen Volume in the Reaction of Methane Combustion

To calculate the volume of oxygen involved in the complete combustion of methane (CH4) in oxygen (O2), we need to use the balanced chemical equation for the reaction:

CH4 + 2O2 → CO2 + 2H2O

From the equation, we can see that one molecule of methane reacts with two molecules of oxygen to produce one molecule of carbon dioxide and two molecules of water.

Given that 3.6 g of water is produced in the reaction, we can use the molar mass of water (H2O) to calculate the number of moles of water produced:

Molar mass of H2O = 2(1.008 g/mol) + 16.00 g/mol = 18.016 g/mol

Number of moles of H2O = mass of H2O / molar mass of H2O = 3.6 g / 18.016 g/mol = 0.1998 mol

Since two moles of water are produced for every two moles of oxygen consumed, we can conclude that the number of moles of oxygen consumed is also 0.1998 mol.

Now, we can use the ideal gas law to calculate the volume of oxygen at standard temperature and pressure (STP). STP is defined as a temperature of 0 degrees Celsius (273.15 K) and a pressure of 1 atmosphere (1 atm).

The ideal gas law equation is:

PV = nRT

Where: P = pressure (in atm) V = volume (in liters) n = number of moles R = ideal gas constant (0.0821 L·atm/(mol·K)) T = temperature (in Kelvin)

At STP, the pressure (P) is 1 atm, the number of moles (n) is 0.1998 mol, and the temperature (T) is 273.15 K.

Substituting these values into the ideal gas law equation, we can solve for the volume (V) of oxygen:

V = nRT / P = (0.1998 mol)(0.0821 L·atm/(mol·K))(273.15 K) / 1 atm = 4.48 L

Therefore, the volume of oxygen involved in the reaction is 4.48 liters.

Please note that the calculation assumes ideal gas behavior and that the reaction occurs at standard temperature and pressure (STP).