# number of moles of KClO3 required to produce 5.6 litres of O2 at STP is-

$2KCl{O}_{3}\to 2KCl+3{O}_{2}\phantom{\rule{0ex}{0ex}}Thus2molesofKCl{O}_{3}gives3×22.4Loxygen\phantom{\rule{0ex}{0ex}}Thatis\phantom{\rule{0ex}{0ex}}67.2Lofoxygenisproducedfrom2moles\phantom{\rule{0ex}{0ex}}So5.6Lofoxygenisproducedfrom=\frac{2}{67.2}×5.6=0.1667moles$

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