Analysis of a KCIO_3-KCI Mixture Aluminum powder can burn in pure oxygen to form

Question

Analysis of a KCIO_3-KCI Mixture Aluminum powder can burn in pure oxygen to form aluminium oxide, Al_2O_3. The balanced equation is: 4 Al_(a) + 3 O_2(g) rightarrow 2 Al_2O_3(s) 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_2O_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 = 27.1 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 KCIO_3? The molar masses are: O_2 = 32.0 g. = 122.6 g. The balanced equation is: 2 KCIL_3(s) MnO_2_rightarrow_Delta 2 KCI(s) + 3 O_2(g)

Explanation / Answer

1)

molAl/molO2      = 4/3

molAl2O3/molAl = 2/4

molO2/molAl2O3 = 3/2

molAl/mol Al2O3 = 4/2

2)

4mol of Al react with 3 moles of O2

4* 27 g     "                3*32

108 g of Al   react with 96 g O2

So answer ,mass of Al = 108 grams

3)

              MnO2, heat

2KClO3(s)      2 KCl(s) + 3 O2 (g)

2 moles KClO3    equivalent to 3 moles O2

2 * 122.6 g KClO3              96 g O2

so O2 produced by 10 g of KClO3 = ( 96/2*122.6 ) * 10 = 3.915 g

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