Question

Люди помогите. Определим объем газа при н.у если сгорает 5л метана

Answer 1

Решение:

CH4+2O2->CO2+2H2O (t) - реакция горения метана.

1). n(CH4)=V(CH4)/Vm, где n - кол-во вещества метана, V- его объем, а Vm -

молярный объем - константа= 22.4л/моль

n(CH4)=5л/22.4л/моль=0.223моль 

2). Согласно уравнению реакции на 1 моль метана приходится один моль газа

(CO2), => n(CO2)=n(CH4)=0.223моль, а поскольку формула для нахождения

объема - V(вещества)=n(вещества)*Vm, а Vm=const, то и их объемы равны.

Ответ: V(CO2)=5л.

Answer 2

CH4 + 2O2 -->CO2 +2H2O
5л

 одинаковые коэффициенты CO2 и CH4
тогда n(CH4)=5/22.4=0.223моль
 n(CH4)=n(CO2)=0.223 MOL
TOGDA  V(CO2)=0.223*22.4=5 л
m(H2O)=2*0.223*18=8.028 г

Answer 3

Calculating Gas Volume at Standard Conditions

To determine the volume of gas at standard conditions when 5 liters of methane is burned, we can use the ideal gas law equation, which is expressed as:

V = (n * R * T) / P

Where: - V = volume of the gas - n = number of moles of the gas - R = ideal gas constant - T = temperature in Kelvin - P = pressure

Given that the gas is at standard conditions (0°C or 273.15 K and 1 atm), we can assume the pressure is 1 atm.

First, we need to calculate the number of moles of methane burned. The balanced chemical equation for the combustion of methane is:

CH4 + 2O2 → CO2 + 2H2O

This equation shows that 1 mole of methane reacts with 2 moles of oxygen to produce 1 mole of carbon dioxide and 2 moles of water.

So, the number of moles of methane (CH4) is equal to the volume of methane (in liters) divided by the molar volume of a gas at standard conditions, which is approximately 22.4 liters per mole.

n(CH4) = Volume of CH4 / Molar volume at STP n(CH4) = 5 L / 22.4 L/mol

Let's calculate the number of moles of methane.