For the reaction 2kclo3(s)→2kcl(s)+3o2(g) calculate how many grams of oxygen form when each quantity of reactant completely reacts.

First, we need to get the molar mass of:

KClO3 = 39.1 + 35.5 + 3*16 = 122.6 g/mol

KCl =39.1 + 35.5 = 74.6 g/mol

O2 = 16*2 = 32 g/mol

From the given equation we can see that:

every 2 moles of KClO3 gives 3 moles of O2

when mass = moles * molar mass

∴ the mass of KClO3 = (2mol of KClO3*122.6g/mol) = 245.2 g

and the mass of O2 then = 3 mol * 32g/mol = 96 g

so, 245.2 g of KClO3 gives 96 g of O2

A) 2.72 g of KClO3: 

when 245.2 KClO3 gives → 96 g  O2

   2.72 g KClO3 gives →  X

X = 2.72 g KClO3 * 96 g O2/245.2 KClO3

    = 1.06 g of O2

B) 0.361 g KClO3:

when 245.2 g KClO3 gives → 96 g O2

     0.361 g KClO3 gives → X

∴ X = 0.361g KClO3 * 96 g / 245.2 g

       = 0.141 g of O2

C) 83.6 Kg KClO3:

when 245.2 g KClO3 gives → 96 g O2

       83.6 Kg KClO3 gives  →  X

∴X = 83.6 Kg* 96 g O2 /245.2 g KClO3

     = 32.7 Kg of O2

D) 22.4 mg of KClO3:

when 245.2 g KClO3 gives → 96 g O2

        22.4 mg KClO3 gives → X

∴X = 22.4 mg * 96 g O2 / 245.2 g KClO3

      = 8.8 mg of O2

     

 

Sorry i dont understand the question can you rephase it?