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diamong [38]
2 years ago
6

Magnesium unites completely and vigorously with oxygen to form magnesium oxide, which contains 60% magnesium by weight. If 1.00

gram of magnesium is sealed in a glass tube with 1.00 gram of oxygen, what will be present in the tube after the reaction has taken place
Chemistry
1 answer:
Dmitry_Shevchenko [17]2 years ago
4 0

Answer:

After the reaction, there will 0.60 g of magnesium oxide and 0.25 g of oxygen gas present in the tube

Explanation:

Equation of the reaction between magnesium and oxygen is given as follows:

2Mg(s) + O₂(g) ---> 2MgO(s)

From the equation of reaction, 2 moles of magnesium reacts with i mole of oxygen gas to produce 1 mole of magnesium oxide

molar mass of magnesium is 24.0 g; molar mass of oxygen gas = 32.0 g; molar mass of magnesium oxide = 40.0 g

Therefore 24 g of magnesium reacts with 32 g of oxygen gas

I.00 g of magnesium will react with (24.0 / 32.0) * 1.00 g of oxygen = 0.75 g of oxygen gas.

Therefore, magnesium is the limiting reagent. Once it is used up, the reaction will stop and the excess oxygen will be left in the tube together with the product, magnesium oxide.

mass of excess oxygen = 1.00 - 0.75 = 0.25 g

mass of magnesium oxide formed = (24.0 / 40.0 g) * 1 = 0.60 g

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Hi there!

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Consider an amphoteric hydroxide, m(oh)2(s), where m is a generic metal. estimate the solubility of m(oh)2 in a solution buffere
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Missing data in your question: (please check the attached photo)
from this balanced equation:
M(OH)2(s) ↔ M2+(aq) + 2OH-(aq) and when we have Ksp = 2x10^-16
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a) SO at PH = 7 
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∴[M2+] =Ksp /[OH]^2
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b) at PH =10
when POH = 14- PH = 14-10 = 4 
when POH = -㏒[OH-]
            4  = -㏒[OH-]
∴[OH] = 1x10^-4 ,by substitution with this value in the Ksp formula
[M2+] = Ksp/ [OH]^2
          = 2x10^-16 / (1x10^-4)^2
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c) at PH= 14 
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[M2+] = Ksp / [OH]^2
          = (2x10^-16) / 1^2
          = 2x10^-16 M


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