Question

Dihydrogen dioxide decomposes into oxygen gas and water. How many moles of Dihydrogen dioxide are required to produce 6.38 moles of oxygen?

Answer 1

12.76 moles of dihydrogen dioxide (H2O2) are required to produce 6.38 moles of oxygen gas (O2), based on the stoichiometry of the balanced decomposition reaction 2 H2O2 → 2 H2O + O2.

Step-by-step explanation:

To determine how many moles of dihydrogen dioxide (H2O2) are required to produce 6.38 moles of oxygen gas (O2), we need to use the stoichiometry of the balanced chemical equation for the decomposition of H2O2. The balanced equation is:

2 H2O2(l) → 2 H2O(l) + O2(g)

From this equation, we see that 2 moles of H2O2 produce 1 mole of O2. To produce 6.38 moles of O2, we would need twice the amount of H2O2, which is:

mol H2O2 = (mol O2) ⋅ (2/1)

This gives us:

6.38 moles O2 ⋅ (2/1) = 12.76 moles H2O2

Therefore, 12.76 moles of dihydrogen dioxide are required to produce 6.38 moles of oxygen gas.

Answer 2

12.76 moles H₂O₂

Explanation

Given:

Moles of oxygen produced = 6.38 moles

What to find:

The moles of Dihydrogen dioxide required to produce 6.38 moles of oxygen.

Step-by-step solution:

Step 1: Write the balanced equation for the decomposition reaction.

2H₂O₂ --------> O₂ + 2H₂O

Step 2: Calculate the moles of H₂O₂ required.

From the balanced equation;

2 mol H₂O₂ produced 1 mol O₂

x mol H₂O₂ is required to produce 6.38 mol O₂

Cross multiply and divide both sides by 1 mol O₂.

`x=\frac{6.38mol\text{ }O₂}{1mol\text{ }O₂}*2mol\text{ }H₂O₂=12.76\text{ }mol\text{ }H₂O₂`

The moles of Dihydrogen dioxide required to produce 6.38 moles of oxygen = 12.76 moles H₂O₂