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prohojiy [21]
3 years ago
13

Your theoretical yield is 81.2 grams, and your actual yield is 78.2 grams. What is the percent yield?

Chemistry
1 answer:
ioda3 years ago
3 0

Answer:

66.2

Explanation:

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For the following reaction, the reactants are favored at equilibrium. Classify each of the reactants and products based on their
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Answer:

See explanation below

Explanation:

First, let's write again the reaction:

CH₃COO⁻ + H(CH₃)₃N⁺  <-----------> CH₃COOH + (CH₃)₃N

Now that the reaction is here, let's remember the basis of the bronsted - lowry theory:

An acid (HA) is a substance that can lose a proton (Hydrogen atom) to form a conjugate base. A base is a substance that accepts the proton (Hydrogen) and form a conjugate acid.

According to this definition, let's see the reaction again.

In the reactants, we see the CH3COO and the H(CH3)N. and the products are CH3COOH and (CH3)3N. The difference? well, we can see that the CH3COO now has a Hydrogen atom, this means that the CH3COO accepted the Hydrogen; this hydrogen was provided by the H(CH3)3N.

Therefore, the acid in this reaction is the H(CH₃)₃N⁺ and the conjugate base will be the (CH₃)₃N

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4 years ago
Determine the enthalpy change for the reaction 2C(s) + 2H2O(g) → CH4(g) + CO2(g) using the following:
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Answer : The enthalpy change for the reaction is, 97.7 kJ

Explanation :

According to Hess’s law of constant heat summation, the heat absorbed or evolved in a given chemical equation is the same whether the process occurs in one step or several steps.

According to this law, the chemical equation can be treated as ordinary algebraic expression and can be added or subtracted to yield the required equation. That means the enthalpy change of the overall reaction is the sum of the enthalpy changes of the intermediate reactions.

The given main chemical reaction is,

2C(s)+3H_2O(g)\rightarrow CH_4(g)+CO_2(g)    \Delta H=?

The intermediate balanced chemical reaction will be,

(1) C(s)+H_2O(g)\rightarrow CO(g)+H_2(g)     \Delta H_1=131.3kJ

(2) CO(g)+H_2O(g)\rightarrow CO_2(g)+H_2(g)    \Delta H_2=41.2kJ

(3) CH_4(g)+H_2O(g)\rightarrow 2H_2(g)+CO(g)    \Delta H_3=206.1kJ

Now we are multiplying reaction 1 by 2 and reversing reaction 3 and then adding all the equations, we get :

(1) 2C(s)+2H_2O(g)\rightarrow 2CO(g)+2H_2(g)     \Delta H_1=2\times 131.3kJ=262.6kJ

(2) CO(g)+H_2O(g)\rightarrow CO_2(g)+H_2(g)    \Delta H_2=41.2kJ

(3) 2H_2(g)+CO(g)\rightarrow CH_4(g)+H_2O(g)    \Delta H_3=-206.1kJ

The expression for enthalpy of main reaction will be,

\Delta H=\Delta H_1+\Delta H_2+\Delta H_3

\Delta H=(262.6)+(41.2)+(-206.1)

\Delta H=97.7kJ

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