Hydrogen peroxide (H2O2) is decomposed according to the following reaction:
2H2O2 (l) ¡÷ 2H2O (l) + O2 (g)
If 6.00 g of aquesous solution containing hydrogen peroxide was decomposed after heating, the mass of oxygen gas collected is 2.40 g, what is the mass percentage of H2O2 in the original solution?
See your CaCO3 problem. Post your work if you get stuck.
To find the mass percentage of H2O2 in the original solution, we need to determine the amount of H2O2 that was decomposed.
First, let's calculate the number of moles of O2 gas produced using the given mass of oxygen gas collected (2.40 g). We can use the molar mass of O2, which is 32 g/mol.
moles of O2 = mass / molar mass = 2.40 g / 32 g/mol = 0.075 mol O2
According to the balanced chemical equation, the stoichiometric ratio between H2O2 and O2 is 2:1. This means that for every 2 moles of H2O2, we get 1 mole of O2.
So, the number of moles of H2O2 decomposed is half of the moles of O2 gas produced.
moles of H2O2 = 0.075 mol O2 / 2 = 0.0375 mol H2O2
Next, we need to determine the molar mass of H2O2, which is 34 g/mol.
Now, we can calculate the mass of H2O2 decomposed using the number of moles of H2O2:
mass of H2O2 decomposed = moles of H2O2 × molar mass of H2O2
= 0.0375 mol × 34 g/mol
= 1.275 g
Since the mass of the original solution is 6.00 g, the mass of H2O2 in the original solution can be calculated by subtracting the mass of H2O2 decomposed from the mass of the original solution:
mass of H2O2 in original solution = mass of original solution - mass of H2O2 decomposed
= 6.00 g - 1.275 g
= 4.725 g
Finally, we can calculate the mass percentage of H2O2 in the original solution:
mass percentage of H2O2 = (mass of H2O2 in original solution / mass of original solution) × 100%
= (4.725 g / 6.00 g) × 100%
= 78.75%
Therefore, the mass percentage of H2O2 in the original solution is 78.75%.