Question

what mass of aluminum oxide is produced from the reaction of 4.63 g of manganese dioxide and 1.07 g of Al? 3MnO2 (s) + 4Al(s) -> 3Mn (s) + 2Al2O3 (s)

Answer 1

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Answer 2

Answer : The correct answer is 2.04 g of Al₂O₃ is produced.

Given : Mass of Al = 1.07 g Mass of MnO₂ = 4.63 g

Mass of Al₂O₃ = ?

Since there are two reactant given , LIMITING REAGENT need to be found .

Limiting Reagent is the that reactant which totally consumed when a reaction completes . The maximum amount of any product can be produced by Limiting Reagent .

Limiting reagent is identified as that reactant which produced less amount of product out of two reactants.

Following steps can be used to calculate mass of aluminum oxide .

Step 1 : To find mass of Al₂O₃ can be produced from Al

Following are the sub steps :

a) To find mole of Al

Mass of Al can be used to find mole of Al by using mole formula as:

$Mole (mol) = \frac{given mass (g)}{atomic mass \frac{g}{mol}}$

Mass of Al = 1.07 g Atomic mass of Al = 26.9 $\frac{g}{mol}$

Plugging value in mole formula :

$Mole = \frac{1.07 g}{26.9 \frac{g}{mol}}$

Mole = 0.040 mol

b) To find mole ratio of Al₂O₃ : Al

Mole ratio is calculated from balanced reaction

Mole of Al₂O₃ in balanced reaction = 2

Mole of Al in balanced reaction = 4

Hence mole ratio of Al₂O₃ : Al = 2 : 4 = 1 : 2

c) To find mole of Al₂O₃

Mole of Al₂O₃ can be calculated using mole of Al and mole ratio

$Mole of Al_2O_3 = 0.040 mol * \frac{1}{2}$

Mole of Al₂O₃ = 0.020 mol

d) To find mass of Al₂O₃

Mass of Al₂O₃ can be find from its mole using mole formula as :

Molar mass of Al₂O₃ = 101.96 $\frac{g}{mol}$

$0.020 mol = \frac{mass of Al_2O_3}{101.96 \frac{g}{mol}}$

Multiplying both side by 101.96 $\frac{g}{mol}$

$0.020 mol * 101.96 \frac{g}{mol} = \frac{mass of Al }{101.96 \frac{g}{mol}} *101.96 \frac{g}{mol}$

Mass of Al₂O₃= 2.04 g

Hence mass of Al₂O₃ produced by 1.07 g of Al = 2.04 g.

Step 2 ) To find mass of Al₂O₃ from MnO₂

Same steps can be used to find mass of Al₂O₃ from mass of MnO₂

a) To find mole of MnO₂

Molar mass of MnO₂ = 86.9$\frac{g}{mol}$

$Mole = \frac{4.63 g}{86.9\frac{g}{mol}}$

Mole = 0.053 mol

b) To find mole ratio of Al₂O₃ : MnO₂

Mole of Al₂O₃ in balanced reaction = 2

Mole of MnO₂ in balanced reaction = 3

Hence mole ratio of Al₂O₃ : MnO₂ = 2 : 3

c) To find mole of Al₂O₃

$Mole of Al_2O_3 = 0.053 mol * \frac{2}{3}$

Mole of Al₂O₃ = 0.035 mol

d) To find mass of Al₂O₃

$0.035 mol = \frac{mass of Al_2O_3}{101.96\frac{g}{mol}}$

Multiplying both side by 101.96 $\frac{g}{mol}$

$0.035 mol * 101.96 \frac{g}{mol} = \frac{mass of Al_2O_3}{101.96 \frac{g}{mol}} *101.96 \frac{g}{mol}$

Mass of Al₂O₃= 3.57 g

Hence mass of Al₂O₃ produced by 4.63 g of MnO₂ = 3.57 g

Step 3 : To find limiting reagent and mass of Al₂O₃ .

Since Al produced less amount of Al2O3 2.04 g than MnO2 which is 3.57 g , so Al is LIMITING REAGENT .

Hence the amount of Al₂O₃ produced by Al will be considered which is 2.04 g.