Question

How many grams of magnesium oxide can be produced if 25.0 g of magnesium are allowed to react with 28 g of oxygen?

2Mg + O2 —> 2MgO

Answer 1

In the reaction of 25.0 g of magnesium with 28 g of oxygen, 41.52 g of magnesium oxide can be produced. Magnesium is the limiting reactant and the amount of product is determined by its molar ratio in the chemical equation.

Step-by-step explanation:

The question asks how many grams of magnesium oxide can be produced from the reaction of 25.0 g of magnesium with 28 g of oxygen. This can be solved using stoichiometric principles.

First, we calculate the molar masses of magnesium (Mg) and oxygen (O2). Magnesium has a molar mass of about 24.31 g/mol, and oxygen (as O2) has a molar mass of about 32.00 g/mol. Using the balanced chemical equation 2Mg + O2 → 2MgO, we can see that 2 moles of Mg (48.62 g) react with 1 mole of O2 (32.00 g) to produce 2 moles of MgO. We need to find the limiting reactant, which is the reactant that will be completely consumed and therefore limit the amount of product formed.

To find the limiting reactant, we divide the mass of each reactant by its respective molar mass to get the moles of each: for magnesium, 25.0 g ÷ 24.31 g/mol ≈ 1.03 moles of Mg; for oxygen, 28 g ÷ 32.00 g/mol ≈ 0.875 moles of O2. Comparing the molar ratios, oxygen is in excess, and magnesium is the limiting reactant.

Therefore, since 1 mole of Mg produces 1 mole of MgO, we can produce 1.03 moles of MgO. The molar mass of MgO is (24.31 + 16.00) g/mol = 40.31 g/mol. Multiplying the moles of MgO by its molar mass gives us: 1.03 moles × 40.31 g/mol ≈ 41.52 g of MgO.

Answer 2

Answer : The mass of
`MgO` produced is, 41.68 grams.

Explanation : Given,

Mass of
`Mg` = 25.0 g

Mass of
`O_2` = 28 g

Molar mass of
`Mg` = 24 g/mol

Molar mass of
`O_2` = 32 g/mol

First we have to calculate the moles of
`Mg` and
`O_2`.

`\text{Moles of }Mg=\frac{\text{Given mass }Mg}{\text{Molar mass }Mg}`

`\text{Moles of }Mg=(25.0g)/(24g/mol)=1.042mol`

and,

`\text{Moles of }O_2=\frac{\text{Given mass }O_2}{\text{Molar mass }O_2}`

`\text{Moles of }O_2=(28g)/(32g/mol)=0.875mol`

Now we have to calculate the limiting and excess reagent.

The balanced chemical equation is:

`2Mg+O_2\rightarrow 2MgO`

From the balanced reaction we conclude that

As, 2 mole of
`Mg` react with 1 mole of
`O_2`

So, 1.042 moles of
`Mg` react with
`(1.042)/(2)=0.521` moles of
`O_2`

From this we conclude that,
`O_2` is an excess reagent because the given moles are greater than the required moles and
`Mg` is a limiting reagent and it limits the formation of product.

Now we have to calculate the moles of
`MgO`

From the reaction, we conclude that

As, 2 mole of
`Mg` react to give 2 mole of
`MgO`

So, 1.042 mole of
`Mg` react to give 1.042 mole of
`MgO`

Now we have to calculate the mass of
`MgO`

`\text{ Mass of }MgO=\text{ Moles of }MgO* \text{ Molar mass of }MgO`

Molar mass of
`MgO` = 40 g/mole

`\text{ Mass of }MgO=(1.042moles)* (40g/mole)=41.68g`

Therefore, the mass of
`MgO` produced is, 41.68 grams.