How many grams of oxygen are required to completely burn out 480 grams of magnesium?
Answer:
First lets have a look at the relation between oxygen and magnesium
So,
2Mg+O2=2MgO
Therefore,from here we can derive the fact that when Magnesium and Oxygen combine their volumes are equal.
Now lets come to the weight of Magnesium you've given.Its 480 grams.We know that the atomic weight of Magnesium is 24grams.So 24*20=480
Hence,in this method we can get the weight of the oxygen required by multiplying it with twenty as well as we know that their volumes in MgO are equal.
As,the atomic weight of Oxygen is 16grams
16*20=320 Grams
Hence 320 grams of oxygen is required to completely burn out 480 grams of magnesium.
:)
The reaction is
2 Mg +O2 >> MgO
Moles Mg = 480 g / 24.312 = 19.7
The ratio between Mg and O2 is 2 : 1
2 : 1 = 19.7 : x
x = 9.87 moles O2 needed
molar mas O2 = 32 g/mol
9.87 mol x 32 g/mol = 316 g of O2
Molar mass of Mg = 24
Molar mass of O2 = 32
2Mg(s) + O2(g) --> MgO(s)
moles of Mg = 480/24 = 20
moles of O2 will be 1/2 x 20 = 10
mass of oxygent = 32 x 10 = 320 gram
Mg reacts with O2 as shown below.
2Mg + O2 = 2MgO
Therefore,
2 gm-atoms of Mg = 1 gram-mole of O2
2*24g of Mg = 32g of O2
48g of Mg = 32g of O2
480g of Mg = ?
= 320g of O2
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