Respuesta :
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:
[tex] Mole (mol) = \frac{given mass (g)}{atomic mass \frac{g}{mol}} [/tex]
Mass of Al = 1.07 g Atomic mass of Al = 26.9 [tex] \frac{g}{mol} [/tex]
Plugging value in mole formula :
[tex] Mole = \frac{1.07 g}{26.9 \frac{g}{mol}} [/tex]
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
[tex] Mole of Al_2O_3 = 0.040 mol * \frac{1}{2} [/tex]
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 [tex] \frac{g}{mol} [/tex]
[tex] 0.020 mol = \frac{mass of Al_2O_3}{101.96 \frac{g}{mol}} [/tex]
Multiplying both side by 101.96 [tex] \frac{g}{mol} [/tex]
[tex] 0.020 mol * 101.96 \frac{g}{mol} = \frac{mass of Al }{101.96 \frac{g}{mol}} *101.96 \frac{g}{mol} [/tex]
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[tex] \frac{g}{mol} [/tex]
[tex] Mole = \frac{4.63 g}{86.9\frac{g}{mol}} [/tex]
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
[tex] Mole of Al_2O_3 = 0.053 mol * \frac{2}{3} [/tex]
Mole of Al₂O₃ = 0.035 mol
d) To find mass of Al₂O₃
[tex] 0.035 mol = \frac{mass of Al_2O_3}{101.96\frac{g}{mol}} [/tex]
Multiplying both side by 101.96 [tex] \frac{g}{mol} [/tex]
[tex] 0.035 mol * 101.96 \frac{g}{mol} = \frac{mass of Al_2O_3}{101.96 \frac{g}{mol}} *101.96 \frac{g}{mol} [/tex]
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.
