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 : 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:

Mass of Al = 1.07 g Atomic mass of Al = 26.9

Plugging value in mole formula :

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 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

Multiplying both side by 101.96

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

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 Al₂O₃ = 0.035 mol

d) To find mass of Al₂O₃

Multiplying both side by 101.96

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.

answer: option (4) is the correct answer.

explanation:

areas which are at low altitudes and near the equator tend to receive direct sunlight from the sun. hence, heat energy or radiations from the sun make these areas warmer and they have warmest water.

whereas areas which are at high altitudes and away from the equator do no receive direct sunlight. hence, no direct contact from the sun makes these regions colder and thus, they have colder water.

thus, we can conclude that the statement areas at low latitudes near the equator have the warmest water because they receive the most direct sunlight best explains the relationship between an area's geography and the temperature of its surface water.