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3 years ago

In the following reaction, how many liters of oxygen will react with 270 liters of ethene (C2H4) at STP?

1 answer:

4 0

According to the equation you have 1 mole of C2H4 and 3 moles of O2.

1 • (22.4L / 270L) = 3 • (22.4L / x)

1/270L = 3/x

x = 3(270) / 1

x = 810 L

810 Liters of oxygen will react with 270 liters of ethene (C2H4) at STP

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3 years ago

What happens to the atomic radius as the atomic number increases?

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3 years ago

El superóxido de potasio, KO2, se emplea en máscaras de respiración para generar

Orlov [11]

Answer:

Reactivo límite: Superóxido de potasio.

Moles de oxígeno producidas: $n_{O_2}=0.11molO_2$

Explanation:

Hola,

En este caso, considerando la reacción química llevada a cabo:

$4KO_2(s) + 2H_2O(l) \rightarrow 4KOH(s) + 3O_2(g)$

Es posible identificar el reactivo límite calculando las moles de superóxido de potasio que serían consumidas por 0.10 mol de agua por medio de la relación molar 4 a 2 que hay entre ellos:

$n_{KO_2}^{consumido\ por\ agua}=0.10molH_2O*\frac{4molKO_2}{2molH_2O} =0.2molKO_2$

Así, dado que solo hay 0.15 mol the superóxido de potasio, podemos decir que este es el reactivo límite. Luego, calculamos las moles de oxígeno producidas, considerando la relación molar 4 a 3 que hay entre el superóxido y el oxígeno:

$n_{O_2}=0.15molKO_2*\frac{3molO_2}{4molKO_2} \\\\n_{O_2}=0.11molO_2$

Best regards.

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3 years ago