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3 years ago

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

Chemistry

1 answer:

Orlov [11]3 years ago

4 0

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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How many grams of a stock solution that is 92.5 percent H2SO4 by mass would be needed to make 250 grams of a 35.0 percent by mas

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94.6 g. You must use 94.6 g of 92.5 % H_2SO_4 to make 250 g of 35.0 % H_2SO_4.

We can use a version of the <em>dilution formula</em>

<em>m</em>_1<em>C</em>_1 = <em>m</em>_2<em>C</em>_2

where

<em>m</em> represents the mass and

<em>C</em> represents the percent concentrations

We can rearrange the formula to get

<em>m</em>_2= <em>m</em>_1 × (<em>C</em>_1/<em>C</em>_2)

<em>m</em>_1 = 250 g; <em>C</em>_1 = 35.0 %

<em>m</em>_2 = ?; _____<em>C</em>_2 = 92.5 %

∴ <em>m</em>_2 = 250 g × (35.0 %/92.5 %) = 94.6 g

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3 years ago

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If 125.0g of nitrogen is reacted with 125.0g of hydrogen, what is the theoretical yield of the reaction? What is the excess reac

MakcuM [25]

Answer:

Hydrogen is the excess reactant

Nitrogen is the limiting reactant

151.6g is theoretical yield

Explanation:

The reaction of N₂ with H₂ to produce NH₃ is:

N₂ + 3H₂ → 2NH₃

To find theoretical yield we need to determine limiting reactant with the moles of each gas as follows:

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125.0g * (1mol / 28g) = 4.46 moles

Hydrogen -Molar mass: 2g/mol-

125.0g * (1mol / 2g) = 62.5 moles of hydrogen

For a complete reaction of 4.46 moles of N2 there are needed:

4.46 moles N2 * (3moles H2 / 1mol N2) = 13.38 moles of hydrogen

As there are 62.5 moles of hydrogen:

<h3>Hydrogen is the excess reactant</h3><h3>Nitrogen is the limiting reactant</h3><h3 />

With nitrogen, the limiting reactant, we determine theoretical moles (Assuming 100% of the reaction occurs) and theoretical yield (In mass):

4.46 moles N2 * (2moles NH3 / 1mol N2) = 8.92 moles of ammonia

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3 years ago

At equilibrium the reactant and product concentrations are constant because a change in one direction is balanced by a change in

Anna [14]

Answer: a. The concentrations of the reactants and products have reached constant values

Explanation:

The reactions which do not go on completion and in which the reactant forms product and the products goes back to the reactants simultaneously are known as equilibrium reactions. For a chemical equilibrium reaction, equilibrium state is achieved when the rate of forward reaction becomes equal to rate of the backward reaction.

Equilibrium state is the state when reactants and products are present but the concentrations does not change with time and are constant.

Equilibrium constant is defined as the ratio of concentration of products to the concentration of reactants each raised to the power their stoichiometric ratios. It is expressed as $K_{eq}$

K is the constant of a certain reaction when it is in equilibrium, while Q is the quotient of activities of products and reactants at any stage other than equilibrium of a reaction.

For a equilibrium reaction,

$A\rightleftharpoons B$

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