Given the following equations:
2 H2O2 (aq) → 2 H2O (l) + O2 (g) and
C2H4 (g) + 3 O2 (g) → 2 CO2 (g) + 2 H2O (l)
The first reaction is the decomposition of hydrogen peroxide into water and oxygen gas. The oxygen gas generated in the first reaction is used in the second (combustion) reaction.
If I have 7.6 g of H2O2, how many grams of CO2 can I produce from the set of reactions?
each two moles of hydrogen peroxide produces one mole of oxygen gas
each three moles of oxygen gas produces two moles of carbon dioxide gas
so the moles of carbon dioxide is one third the starting moles of hydrogen peroxide
use the molar masses to find the amount produced
50.0 g of water are decomposed. 25.5L of Oxygen gas are produced as a result. What is the percent yield according to the following reaction. H2O --> H2 + O2 *
First, we need to determine the molar mass of H2O2. Hydrogen (H) has a molar mass of approximately 1.01 g/mol, and oxygen (O) has a molar mass of approximately 16.00 g/mol. Since there are 2 hydrogen atoms and 2 oxygen atoms in H2O2, its molar mass is:
(2 * 1.01 g/mol) + (2 * 16.00 g/mol) = 34.02 g/mol.
Next, we can use the molar ratio between H2O2 and CO2 from the balanced equations to find how many moles of CO2 can be produced. From the first equation, we know that 2 moles of H2O2 produce 2 moles of H2O and 1 mole of O2. From the second equation, we know that 1 mole of C2H4 reacts with 3 moles of O2 to produce 2 moles of CO2 and 2 moles of H2O.
Since the oxygen necessary for the combustion reaction is derived from the decomposition reaction, we need to find how many moles of H2O2 are required to generate the number of moles of O2 needed for the second reaction.
From the first equation, we have a 1:1 mole ratio between H2O2 and O2. So, the number of moles of O2 produced from 7.6 g of H2O2 can be calculated using the molar mass of H2O2:
Number of moles of H2O2 = Mass of H2O2 / Molar mass of H2O2
= 7.6 g / 34.02 g/mol ≈ 0.2238 mol
Since the mole ratio between O2 and H2O2 is 1:1, we also have 0.2238 mol of O2.
Now, we use the mole ratio between O2 and CO2 from the second equation to find how many moles of CO2 can be produced:
Number of moles of CO2 = (Number of moles of O2) * (2 moles of CO2 / 1 mole of O2)
= 0.2238 mol * (2/1) ≈ 0.4476 mol
Finally, we can calculate the mass of CO2 produced using the molar mass of CO2:
Mass of CO2 = Number of moles of CO2 * Molar mass of CO2
= 0.4476 mol * (2 * 12.01 g/mol + 16.00 g/mol)
≈ 17.0326 g
Therefore, you can produce approximately 17.03 grams of CO2 from 7.6 grams of H2O2 according to the given set of reactions.
To find out how many grams of CO2 can be produced from the given reactions, we need to follow a few steps:
Step 1: Calculate the number of moles of H2O2.
The molar mass of H2O2 is 34.0147 g/mol.
The number of moles of H2O2 can be calculated using the formula:
moles = mass / molar mass
moles of H2O2 = 7.6 g / 34.0147 g/mol
Step 2: Use the stoichiometry of the reaction to determine the mole ratio between H2O2 and CO2.
From the balanced equation:
2 H2O2 (aq) → 2 H2O (l) + O2 (g)
C2H4 (g) + 3 O2 (g) → 2 CO2 (g) + 2 H2O (l)
The mole ratio of H2O2 to CO2 is 2:2 or 1:1.
Step 3: Calculate the number of moles of CO2 that can be produced.
Since the mole ratio of H2O2 to CO2 is 1:1, the moles of CO2 will be the same as the moles of H2O2.
moles of CO2 = moles of H2O2
Step 4: Calculate the mass of CO2.
The molar mass of CO2 is 44.0095 g/mol.
The mass of CO2 can be calculated using the formula:
mass = moles * molar mass
mass of CO2 = moles of CO2 * molar mass of CO2
By plugging in the value of moles of H2O2 calculated in Step 1, we can find the mass of CO2.
mass of CO2 = moles of H2O2 * molar mass of CO2
Now, let's calculate the mass of CO2:
moles of H2O2 = 7.6 g / 34.0147 g/mol
moles of CO2 = moles of H2O2
molar mass of CO2 = 44.0095 g/mol
mass of CO2 = moles of CO2 * molar mass of CO2
By substituting the values in the formula, you can find the mass of CO2 that can be produced.