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Mars2501 [29]
3 years ago
13

The reducing agent in this catalytic hydrogenation reaction was molecular hydrogen (H2), which was produced in situ (in the reac

tion mixture) during the course of the reaction. a) During the course of the experiment, how did you know that H2 had been formed? b) Write a balanced chemical equation for the reaction that produced H2. c) Was H2 the limiting reactant? Why or why not? Show the necessary calculations to support your answer.
Chemistry
1 answer:
creativ13 [48]3 years ago
4 0

Answer:

a)  After the balloon inflated after 440 uL of dropwise due to the reaction of 1-Decene and the solution in the conical vial. b) 4NaBH_{4} + 2HCl + 7H_{2}O ⇒ 16H_{2(g)}+ 2NaCl + Na_{2}B_{4}O_{7} c) No H_{2} was not the limiting reactant.

Explanation:

Generally, hydrogenation is the chemical reaction between a compound or element and molecular hydrogen in the presence of catalysts such as platinum.

a) After the balloon inflated after 440 uL of dropwise 1-Decene solution was added due to the reaction between 1-Decene and the solution in the conical vial.

b)  4NaBH_{4} + 2HCl + 7H_{2}O ⇒ 16H_{2(g)}+ 2NaCl + Na_{2}B_{4}O_{7}

c) H_{2} was not the limiting reactant based on the mol to mol ratio of H_{2} and decane which is 1:1. Therefore, if 0.8 mol of decane was produced then 0.8 mol of H_{2} would also be produced.

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At a given temperature the vapor pressures of hexane and octane are 183 mmhg and 59.2 mmhg, respectively. Calculate the total va
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Total vapor pressure can be calculated using partial vapor pressures and mole fraction as follows:

P=X_{A}P_{A}+X_{B}P_{B}

Here, X_{A} is mole fraction of A, X_{B} is mole fraction of B, P_{A} is partial pressure of A and P_{B} is partial pressure of B.

The mole fraction of A and B are related to each other as follows:

X_{A}+X_{B}=1

In this problem, A is hexane and B is octane, mole fraction of hexane is given 0.580 thus, mole fraction of octane can be calculated as follows:

X_{octane}=1-X_{hexane}=1-0.58=0.42

Partial pressure of hexane and octane is given 183 mmHg and 59.2 mmHg respectively.

Now, vapor pressure can be calculated as follows:

P=X_{hexane}P_{hexane}+X_{octane}P_{octane}

Putting the values,

P=(0.580)(183 mmHg)+(0.420)(59.2 mmHg)=131 mmHg

Therefore, total vapor pressure over the solution of hexane and octane is 131 mmHg.

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