7A. Hydrogen peroxide decomposes into water vapor and oxygen. Suppose you have 225 g of a solution that is 50% hydrogen peroxide
by mass. What volume of oxygen can be produced at 25oC and 792 torr?
2 answers:
Answer: 38.7L
Explanation:
First we must write the balanced reaction equation. We can obtain the mass of H2O2 in the solution by applying the fact that it accounts for 50% by mass of solution as shown in the image attached. Then we calculate the number of moles of hydrogen peroxide present. From PV=nRT, we obtain the volume of H2O2. By applying the stoichiometry of the reaction, we obtain the volume of oxygen evolved in the reaction as shown in the image attached
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Answer:
Explanation:
According to the Arrhenius equation,
or,
where,
= rate constant at
=
= rate constant at
=
= activation energy for the reaction = 262 kJ/mol = 262000J/mol
R = gas constant = 8.314 J/mole.K
= initial temperature =
= final temperature =
Now put all the given values in this formula, we get
Therefore, the value of the rate constant at 775.0 K is
Answer:
Explanation:
Hello,
In this case, the described chemical reaction is:
Thus, for the given reacting masses, we must identify the limiting reactant for us to determine the maximum mass of water that could be produced, therefore, we proceed to compute the available moles of ethane:
Next, we compute the moles of ethane consumed by 13.0 grams of oxygen by using the 1:7/2 molar ratio between them:
Thus, we notice there are less available moles of ethane, for that reason, it is the limiting reactant, thereby, the maximum amount of water is computed by considering the 1:3 molar ratio between ethane and water:
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