Respuesta :
Decomposition of a potassium superoxide happens according to the scheme:
4KO₂(s) + 2H₂O(l) → 4KOH(aq) + 3O₂(g)
m(K₂O)=4M(K₂O)m(O₂)/{3M(O₂)}
m(K₂O)=4×71.1×6.5/{3×32.0}≈19.3 g
4KO₂(s) + 2H₂O(l) → 4KOH(aq) + 3O₂(g)
m(K₂O)=4M(K₂O)m(O₂)/{3M(O₂)}
m(K₂O)=4×71.1×6.5/{3×32.0}≈19.3 g
Answer: The mass of [tex]KO_2[/tex] needed is 19.3 grams.
Explanation:
To calculate the number of moles, we use the equation:
[tex]\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}[/tex] .....(1)
Given mass of oxygen gas = 6.5 g
Molar mass of oxygen gas = 32 g/mol
Putting values in equation 1, we get:
[tex]\text{Moles of oxygen gas}=\frac{6.5g}{32g/mol}=0.203mol[/tex]
The chemical equation follows:
[tex]4KO_2+2CO_2\rightarrow 2K_2CO_3+3O_2[/tex]
By Stoichiometry of the reaction:
3 moles of oxygen gas is produced by 4 moles of [tex]KO_2[/tex]
So, 0.203 moles of oxygen gas is produced by = [tex]\frac{4}{3}\times 0.203=0.271mol[/tex] of [tex]KO_2[/tex]
Now, calculating the mass of [tex]KO_2[/tex] by using equation 1, we get:
Molar mass of [tex]KO_2[/tex] = 71.1 g/mol
Moles of [tex]KO_2[/tex] = 0.271 moles
Putting values in equation 1, we get:
[tex]0.271mol=\frac{\text{Mass of }KO_2}{71.1g/mol}\\\\\text{Mass of }KO_2=(0.271mol\times 71.1g/mol)=19.3g[/tex]
Hence, the mass of [tex]KO_2[/tex] needed is 19.3 grams.
