Question #c6feb

1 Answer
Mar 4, 2017

Answer:

#"0.88123990202g of O2"#

Explanation:

First convert 4.5g of #KClO_3# to moles

#"Moles"# = #"weight in grams"/"molar mass"#

Molar mass of KClO3 = 122.55 g/mol

#"4.5g"/"122.55g/mol" = 0.03671970624mol#

Use mole ratio of reaction i.e

#2KClO_3 : 2KCl : 3O_2#

If 2 = 0.03671970624mol then 3 =

#"0.03671970624mol"/2 * 3 = "0.05507955936mol of O2"#

Weight of 0.05507955936mol of #O_2#

#Weight = "Molar mass" xx "moles"#

Molar mass of #O_2# = 15.9994g/mol

#"15.9994g/mol" xx "0.05507955936mol" = "0.88123990202g of O2"#