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

what website did you find this paper at?

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) = (given mass (g))/(atomic mass (g)/(mol))`

Mass of Al = 1.07 g Atomic mass of Al = 26.9
`(g)/(mol)`

Plugging value in mole formula :

`Mole = (1.07 g)/(26.9 (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 * (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
`(g)/(mol)`

`0.020 mol = (mass of Al_2O_3)/(101.96 (g)/(mol))`

Multiplying both side by 101.96
`(g)/(mol)`

`0.020 mol * 101.96 (g)/(mol) = (mass of Al )/(101.96 (g)/(mol)) *101.96 (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
`(g)/(mol)`

`Mole = (4.63 g)/(86.9(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 * (2)/(3)`

Mole of Al₂O₃ = 0.035 mol

d) To find mass of Al₂O₃

`0.035 mol = (mass of Al_2O_3)/(101.96(g)/(mol))`

Multiplying both side by 101.96
`(g)/(mol)`

`0.035 mol * 101.96 (g)/(mol) = (mass of Al_2O_3)/(101.96 (g)/(mol)) *101.96 (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.