Respuesta :
Answer:
16.84 dm3 of hydrogen peroxide will be used to produce 8.42 dm3 of oxygen.
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.
Answer:
0.01596 M
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