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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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Answer:

n_{Mg}=3.50molMg\\\\ n_{Cl}=7.00molCl\\\\n_O=28.0molO

Explanation:

Hello there!

In this case, according to the given information it turns out possible for us to realize that one mole of the given compound, Mg(ClO₄)₂, has one mole of Mg, two moles of Cl and eight moles of O; thus, we proceed as follows:

n_{Mg}=3.50molMg(ClO_4)_2*\frac{1molMg}{1molMg(ClO_4)_2}=3.50molMg\\\\ n_{Cl}=3.50molMg(ClO_4)_2*\frac{2molCl}{1molMg(ClO_4)_2}=7.00molCl\\\\n_O=3.50molMg(ClO_4)_2*\frac{8molO}{1molMg(ClO_4)_2}=28.0molO

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A sample of an ideal gas at 1.00 atm and a volume of 1.84 L was placed in a weighted balloon and dropped into the ocean. As the
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Since we are dealing with an ideal gas, we can calculate the final volume using Boyle's law.

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