3 QUESTIONS A 4.0-mol sample of KClO 3 was decomposed according to the equation

Question

3 QUESTIONS

A 4.0-mol sample of KClO3 was decomposed according to the equation
2KClO3(s) --> 2KCl(s) + 3O2(g)
How many moles of O2 are formed assuming 100% yield? (Points : 2)        2.0 mol
       2.5 mol
       3.0 mol
       6.0 mol
       4.5 mol

The rusting of iron is represented by the equation 4Fe + 3O2 --> 2Fe2O3. If you have a 2.00-mol sample of iron, how many moles of Fe2O3 will there be after the iron has rusted completely? (Points : 2)        0.50 mol
       0.75 mol
       1.0 mol
       1.50 mol
       2.0 mol

Explanation / Answer

1) The number of mols of O2 formed with 100% yield = (4.0 mol)(3 mol O2 / 2 mol KClO3) = 6 mol O2 (option C)

2) The number of mols of Fe2O3 formed = (2.00 mol Fe)((2 mol Fe2O3 / 4 mol Fe) = 1 mol Fe2O3

3) The number of mols of NH3 would be formed = (1 mol H2)(2 mol NH3 / 3 mol H2) = 0.667 mol NH3

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