Analysis of a KClO_3-KCl Mixture Aluminum powder can burn in pure oxygen to form

Question

Analysis of a KClO_3-KCl Mixture Aluminum powder can burn in pure oxygen to form aluminum oxide Al_2O_3. The balanced equation is: 4 Al_ + 3 O_2(g) rightarrow 2 Al_2O_ (s) Write numbers in the small blanks to identify four male ratios that exist in this balanced equation. You may want to refer to the Chemical Arithmetic - Equations exercises: 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_1 = 32.0 g. Please you're your calculations below. Masses of Al = ______ How many grams of oxygen, O_2, would be produced in the complete decomposition of 10.0 g of KClO_3? The molar masses are: O_2 = 32.0 g, KClO = 122.6 g. The balanced equation is 2 KClO_3 2 KCl_ + 3 O_2

Explanation / Answer

1. For the given reaction,

mol ratios would be,

4 mol Al/3 mol O2           1 mol Al2O3/2 mol Al           3 mol O2/2 mol Al2O3         2 mol Al/1 mol Al2O3

2. moles of O2 reacted = 96 g/32 g/mol

                                     = 3 mol

So,

mass of Al needed = 4 x 3 x 27/3

                               = 108 g

3. moles of KClO3 reacted = 10 g/122.6 g/mol

                                           = 0.0816 mol

So,

mass of O2 produced = 3 x 0.0816 x 32/2

                                    = 3.92 g

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