A hot solution of potassium persulfate slowly decomposes, giving oxygen as one of the products as;

2K2S2O8+2H2O-->4KHSO4+O2

Calculate the maximum volume of oxygen gas that can be produced at 80C when a solution containing 0.100 mol of potassium persulfate decomposes as shown above.

[At 80 °C 1 mol of oxygen has a volume of 29.0 dm3]

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`2K_2S_2O_8+ 2H_2Orarr4KHSO_4+O_2`

Mole ratio

`2K_2S_2O_8:O_2 = 2:1`

Amount of `2K_2S_2O_8` decomposed `= 0.1 mol`

Amount of `O_2` formed `= 0.1/2 = 0.05 mol`

It is given that the volume of 1 mol of `O_2` at 80C is `29dm^3` .

Volume of 0.05 moles of `O_2` at 80C `= 29xx0.05 = 1.45dm^3`

*So the maximum volume of `O_2` that can be obtained by decomposing 0.1 mol of `2K_2S_2O_8 ` is `1.45dm^3` .*

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