Question

How many grams of ko2 are needed to form 6.5 g of o2?

Answer 1

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

Answer 2

Answer: The mass of $KO_2$ needed is 19.3 grams.

Explanation:

To calculate the number of moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$      .....(1)

Given mass of oxygen gas = 6.5 g

Molar mass of oxygen gas = 32 g/mol

Putting values in equation 1, we get:

$\text{Moles of oxygen gas}=\frac{6.5g}{32g/mol}=0.203mol$

The chemical equation follows:

$4KO_2+2CO_2\rightarrow 2K_2CO_3+3O_2$

By Stoichiometry of the reaction:

3 moles of oxygen gas is produced by 4 moles of $KO_2$

So, 0.203 moles of oxygen gas is produced by = $\frac{4}{3}\times 0.203=0.271mol$ of $KO_2$

Now, calculating the mass of $KO_2$ by using equation 1, we get:

Molar mass of $KO_2$ = 71.1 g/mol

Moles of $KO_2$ = 0.271 moles

Putting values in equation 1, we get:

$0.271mol=\frac{\text{Mass of }KO_2}{71.1g/mol}\\\\\text{Mass of }KO_2=(0.271mol\times 71.1g/mol)=19.3g$

Hence, the mass of $KO_2$ needed is 19.3 grams.