The following equation is one way to prepare oxygen in a lab.2kclo3 → 2kcl + 3o2molar mass info: mm o2 = 32 g/mol mm kcl = 74.55 g/mol mm kclo3 = 122.55 g/molif 25.6 g of kcl are produced, then how many moles of kclo3 were consumed? a. 0.209 molb. 0.687 molc. 0.343 mold. 0.172 mol

Molar mass:

KCl = 74.55 g/mol

KClO3 = 122. 55 g/mol

  Calculation of the mass of KClO3 :

2 KClO3 =  2 KCl + 3 O2

2* 122.55 g KClO3 2 * 74.55 g KCl
mass KClO3 ?? 25.6 g KCl

mass KClO3 = 25.6 * 2 * 122.55 / 2 * 74.55

mass KClO3 = 6274.56 / 149.1

mass = 42.082 g of  KClO3

Therefore:

1 mole KClO3 122.55 g
?? moles KClO3 42.082 g

moles KClO3 = 42.082 * 1 / 122.55

moles KClO3 = 42.082 / 122.55

=> 0.343 moles of KClO3

Answer C

hope this helps!