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`18.625g KCl` is formed due to decomposition of `KClO_(4)` in reaction `KClO_(4)(s)rarr KCl(s)+2O_(2)(g)` Find volume of `O_(2)` obtained at `STP`[Atomic mass `K = 39 Cl = 35.5]` |
Answer» `{:(KClO_(4)(s),rarr,KCl(s),+,2O_(2)(g)),(,,(18.625)/(74.5)mol,,(2xx18.626)/(74.5)),(,,=0.25,,=0.5mol):}` Vol of `O_(2)` at `STP = 0.5 xx 22.4 = 11.2L` . |
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