Question

PLEASE HELP! CHEMISTRY STOICIOMETRY QUESTION!

A 47L bottle of hydrogen peroxide will eventually decomposes into water and oxygen at SATP if the lid is left open, what concentration of hydrogen peroxide is required to react 8.42L of oxygen?

Answer 1

0.01596 M

Step-by-step explanation:

This is a stoichiometry problem hence we need to obtain the balanced reaction equation before we can effectively proceed in solving the problem at hand.

Given the reaction equation;

2H2O2(aq) -------> 2H2O(l) + O2(g)

2 moles of hydrogen peroxide yields 1 mole of oxygen

I mole of oxygen occupies 22.4 L volume.

Hence

2 moles of hydrogen peroxide produces 22.4 L of oxygen

x moles of hydrogen peroxide produces 8.42 L of oxygen

x= 2 × 8.42/22.4

x= 0.75 moles of hydrogen peroxide

From

Number of moles = concentration × volume

Number of moles= 0.75 moles

Volume = 47 L

Concentration of hydrogen peroxide= number of moles/ volume

Concentration of hydrogen peroxide= 0.75 moles/47 L

Concentration of hydrogen peroxide= 0.01596 M

Answer 2

16.84 dm3 of hydrogen peroxide will be used to produce 8.42 dm3 of oxygen.

Step-by-step explanation:

Equation for the reaction:

H2O2 --------> 1/2 O2 + H20

1 mole of hydrogen peroxide reacts to form half mole of oxygen

At STP, 22.4 dm3 of hydrogen peroxide will react to form 1/2* 22.4 dm3 of oxygen

So 22.4 dm3 of hydrogen peroxide forms 11.2 dm3 of oxygen

If 8.42L of oxygen were formed, how many volume of hydrogen peroxide is used?

From 22.4 dm3 of H2O2 = 11. 4 dm3 of O2

(22.4 * 8.42 / 11.4 )dm3 of hydrogen peroxide will form 8.42 dm3 of oxygen.

= 16.84 dm3 of hydrogen peroxide will be used.

In the reaction of hydrogen peroxide to form oxygen, 16.84 dm3 of hydrogen peroxide will give off 8.42 dm3 of oxygen.