Question

V(CH4)=0,8 л V(C2H2) -?

Answer 1

Дано:

V(CH₄) = 0,8 л

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Найти:

V(C₂H₂) - ?

Решение:

1) Для начала мы запишем уравнение реакций и метана в ацетилен:

2CH₄ (t°C) → C₂H₂ + 3H₂↑

2) Для начала найдем количества вещества у метана, а потом у ацетилена:

Vm = 22,4 л/моль

n(CH₄) = V(CH₄)/Vm = 0,8 л / 22,4 л/моль ≈ 0,036 моль ⇒ n(C₂H₂) = 1/2 × n(CH₄) = 1/2 × 0,036 моль = 0,018 моль

3) Далее находим объем ацетилена:

V(C₂H₂) = n(C₂H₂)×Vm = 0,018 моль × 22,4 л/моль = 0,4032 л ≈ 0,4 л

Ответ: V(C₂H₂) = 0,4 л

$DK954$

Answer 2

To determine the volume of acetylene gas (C2H2) produced when methane gas (CH4) is completely reacted, we can use the stoichiometry of the balanced chemical equation for the combustion of methane:

CH4(g) + 2O2(g) -> CO2(g) + 2H2O(g)

From the balanced equation, we can see that one mole of methane (CH4) produces one mole of carbon dioxide (CO2) and two moles of water (H2O). This means that the volume ratio of CH4 to CO2 to H2O is 1:1:2.

Given that the initial volume of methane (V(CH4)) is 0.8 liters, when it completely reacts, it will produce the same volume of carbon dioxide (V(CO2)) and twice that volume of water vapor (V(H2O)).

So, V(CO2) = 0.8 liters (since it's a 1:1 ratio with CH4), and V(H2O) = 2 * 0.8 liters = 1.6 liters (since it's a 2:1 ratio with CH4).

Now, to find the volume of acetylene gas (C2H2), we need to use the balanced equation for the combustion of acetylene:

2C2H2(g) + 5O2(g) -> 4CO2(g) + 2H2O(g)

From this equation, we can see that two moles of acetylene (C2H2) produce four moles of carbon dioxide (CO2) and two moles of water (H2O). This means that the volume ratio of C2H2 to CO2 to H2O is 2:4:2, which simplifies to 1:2:1.

So, V(C2H2) will be half the volume of V(CO2).

V(C2H2) = 0.5 * V(CO2) = 0.5 * 0.8 liters = 0.4 liters

Therefore, when 0.8 liters of methane (CH4) completely reacts, it will produce 0.4 liters of acetylene gas (C2H2).