Aluminum powder can burn in pure oxygen to form aluminum oxide, Al_2O_3, The bal

Question

Aluminum powder can burn in pure oxygen to form aluminum oxide, Al_2O_3, The balanced equation is: 4 Al_(a) + 3 O_2 (g) rightarrow 2 Al_2 O_3 (a) Write numbers in the small blanks to identify four mole ratios that exist in this balanced equation. You may want to refer to the Chemical Arithmetic -Equations exercises: __ mol Al/mol O_2 __ mol Al_2 O_3/__ mol Al __ mol O_2/__ mol Al_2 O_3 __mol Al/__ mol Al_2 O_3 Referring to the balanced equation in question 1, what mass of aluminum is required to just completely react with 96.0 g of O_2? The molar masses are: Al = 27.0 g, O_2 = 32.0 g. Please show your calculations below. Mass of Al = How many grams of oxygen, O_2, would be produced in the complete decomposition of 10.0 g of KClO_2? The molar masses are: O_2 = 32.0 g. KClO_3 = 122.6 g. The balanced equation is: 2 KClO_3 rightarrow Mn O_2 Delta 2 KCl_(s) + 3 O_2 (g)

Explanation / Answer

Question 1.

4 mol of Al --> 3 mol of O2 so --> 4/3

2 mol of Al2O3 --> 4 mol of Al ---> 2/4

3 mol of O2 --> 2 mol of Al2O3 --> 3/2

4 mol of Al --> 2 mol of Al2O3 --> 4/2

Q2.

from this balanc

mass of Al required fo

m = 96 g of O2

mol of O2 = mass/MW = 96/32 = 3 mol of O2

from our ratios..

3 mol of O2 requires --> 4 mol of Al

mass = mol*MW = 4*27 = 108 g of Aluminium required

Q3

mass of O" for:

m = 10 g of KClO3

mol KClO3 = mass/MW = 10/(122.6) = 0.08156 mols

2 mol of KClO3 = 3 mol of O2

then

0.08156 mol --> 1.5*0.08156 = 0.12234 mol of O2 will be produces

mass = mol*MW = 0.12234*32 = 3.91488 g of O2

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