2kmno₄ → k₂mno₄ + mno₂ + o₂\nhow many grams of oxygen form from 7.0 moles of kmno₄?\nstep 1: how many moles…

Answer:

56 g

Explanation:

Step1: Determine mole - ratio

From the balanced equation $2KMnO_4\rightarrow K_2MnO_4 + MnO_2+O_2$, the mole - ratio of $KMnO_4$ to $O_2$ is 2:1.

Step2: Calculate moles of $O_2$

If we have 7.0 moles of $KMnO_4$, the moles of $O_2$ formed is $n_{O_2}=\frac{7.0\ mol\ KMnO_4\times1\ mol\ O_2}{2\ mol\ KMnO_4}=3.5\ mol$.

Step3: Calculate mass of $O_2$

The molar mass of $O_2$ is $M = 32\ g/mol$. Using the formula $m = n\times M$, where $n = 3.5\ mol$ and $M=32\ g/mol$, we get $m = 3.5\ mol\times32\ g/mol = 56\ g$.