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Answer 1 · 2017-12-31T22:22:08

$= 31.0 m L of O_{2}$ $=31.0mL " of " O_2$

Write and balance the equation
$2 K C l O \to 2 K C l + O_{2}$ $2KClO->2KCl+O_2$
I assumed that the $m K C l O = 0.250 g$ $mKClO=0.250g$ . Per convention, convert it to moles. The molar mass of $K C l O = \frac{90.5 g}{m o l}$ $KClO=(90.5g)/(mol)$ which is obtainable from the periodic table; i.e.,
$etaKClO=0.250cancel(gKClO)xx(1molKClO)/(90.5cancel(gKClO)$
$etaKClO=0.002762mol$
Referring to the balanced equation, find the $etaO_2$ , where:
$=0.002762cancel(molKClO)xx(1molO_2)/(2cancel(molKClO))$
$=0.001381molO_2$
At $STP$ , mol of a gas will occupy a volume of 22.4L. This relationship will serve as the conversion factor to find the volume of $O_2$ collected at this condition.
$V_(O_2)=0.001381cancel(molO_2)xx(22.4L" of " O_2)/(1cancel(molO_2))$
$V_(O_2)=0.03094L" of " O_2~~31.0mL" of "O_2$