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Sagot :
When 25.0 g of Mg react with 25.0 g of O₂ it produce 41.9g of MgO.
Chemical reaction:
2Mg + O₂ → 2 MgO
Given,
Mass of Mg = 24g
Mass of O₂ = 25 g
Molar mass of O₂ = 16×2
= 32 g
Molar mass of Mg = 24.3
Moles of Mg =25/24.3
=1.04mole
Moles of O₂ =25/32
=0.78mole
From above reaction we can say that 2moles of mg react with 1moles of O₂ to give 2moles of MgO.
So, 1 mole of Mg react with 0.5 mole of O₂ to produce 1mole of MgO.
So, 1.04 react with= 0.5× 1.04
= 0.520 g oxygen to give 1.04 moles of MgO.
From above we can say that Mg is limiting reagent.
Moles of MgO =1.04
Molar mass of MgO =40.3
Mole =given mass/ molar mass
1.04 =mass/ 40.3
Mass = 1.04×40.3
41.9 g
Thus, from above we concluded that 25.0 g of Mg react with 25.0 g of O₂ it produce 41.9g of MgO.
Learn more about limiting reagent: https://brainly.com/question/26905271
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