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1 year ago

A 10.0-g sample of magnesium reacts with oxygen to form 16.6 g of magnesium oxide. How many grams of oxygen reacted?

Chemistry

1 answer:

kodGreya [7K]1 year ago

3 0

6.6 grams of oxygen reacted with magnesium to give magnesium oxide.

<h3>What is law of conservation of mass?</h3>

When a chemical reaction takes place the total mass of reactants is always equal to the mass of products.As in a chemical reaction the atoms are neither created nor destroyed rather they rearrange themselves.

The reaction between magnesium and oxygen is as follows,

2 Mg+O₂ $\rightarrow$ 2 MgO

According to the question, 10 g of magnesium is utilized to produce 16.6 g of magnesium oxide.In order to find mass of oxygen utilized ,

mass of oxygen=mass of magnesium oxide-mass of magnesium

substituting the given data,

mass of oxygen=16.6-10=6.6 g

Thus, 6.6 grams oxygen reacted to produce 16.6 grams of magnesium oxide.

Learn more about law of conservation of mass ,here:

brainly.com/question/13383562

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Read 2 more answers

What is the limiting reagent when a 2.00 g sample of ammonia is mixed with 4.00 g of oxygen?

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Answer:

Ammonia is limiting reactant

Amount of oxygen left = 0.035 mol

Explanation:

Masa of ammonia = 2.00 g

Mass of oxygen = 4.00 g

Which is limiting reactant = ?

Balance chemical equation:

4NH₃ + 3O₂ → 2N₂ + 6H₂O

Number of moles of ammonia:

Number of moles = mass/molar mass

Number of moles = 2.00 g/ 17 g/mol

Number of moles = 0.12 mol

Number of moles of oxygen:

Number of moles = mass/molar mass

Number of moles = 4.00 g/ 32 g/mol

Number of moles = 0.125 mol

Now we will compare the moles of ammonia and oxygen with water and nitrogen.

NH₃ : N₂

4 : 2

0.12 : 2/4×0.12 = 0.06

NH₃ : H₂O

4 : 6

0.12 : 6/4×0.12 = 0.18

O₂ : N₂

3 : 2

0.125 : 2/3×0.125 = 0.08

O₂ : H₂O

3 : 6

0.125 : 6/3×0.125 = 0.25

The number of moles of water and nitrogen formed by ammonia are less thus ammonia will be limiting reactant.

Amount of oxygen left:

NH₃ : O₂

4 : 3

0.12 : 3/4×0.12= 0.09

Amount of oxygen react = 0.09 mol

Amount of oxygen left = 0.125 - 0.09 = 0.035 mol

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