Question

Mg also burns in carbon dioxide. In this case, Mg reacts with CO₂ to produce MgO and solid Carbon (C(s)). How many grams of CO₂ are needed to consume 1.00 g of Mg

Answer 1

In the reaction 2Mg(s) + CO₂(g) → 2MgO(s) + C(s), approximately 0.904 grams of CO₂ are needed to consume 1.00 g of Mg.

Step-by-step explanation:

In the reaction 2Mg(s) + CO₂(g) → 2MgO(s) + C(s), 2 moles of Mg reacts with 1 mole of CO₂ to produce 2 moles of MgO and 1 mole of solid Carbon (C(s)). Since the molar mass of Mg is 24.31 g/mol and the molar mass of CO₂ is 44.01 g/mol, we can use stoichiometry to calculate the grams of CO₂ needed to consume 1.00 g of Mg.

First, calculate the moles of Mg in 1.00 g using its molar mass:

1. Moles of Mg = Mass of Mg / Molar mass of Mg
2. Moles of Mg = 1.00 g / 24.31 g/mol
3. Moles of Mg = 0.0411 mol

Since the mole ratio between Mg and CO₂ is 2:1, we can use a proportion to calculate the moles of CO₂:

1. Moles of CO₂ / Moles of Mg = Molar ratio
2. Moles of CO₂ / 0.0411 mol = 1 mol / 2 mol
3. Moles of CO₂ = 0.02055 mol

Finally, we can convert the moles of CO₂ to grams using its molar mass:

1. Mass of CO₂ = Moles of CO₂ x Molar mass of CO₂
2. Mass of CO₂ = 0.02055 mol x 44.01 g/mol
3. Mass of CO₂ = 0.9038 g

Therefore, approximately 0.904 grams of CO₂ are needed to consume 1.00 g of Mg.

Answer 2

Answer: The mass of
`CO_2` needed is, 0.915 grams.

Explanation : Given,

Mass of
`Mg` = 1.00 g

Molar mass of
`Mg` = 24 g/mol

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

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

`\text{Moles of }Mg=(1.00g)/(24g/mol)=0.0417mol`

Now we have to calculate the moles of
`CO_2`

The balanced chemical equation is:

`2Mg+CO_2\rightarrow 2MgO+C`

From the balanced reaction we conclude that

As, 2 mole of
`Mg` react with 1 mole of
`CO_2`

So, 0.0417 moles of
`Mg` react with
`(0.0417)/(2)=0.0208` moles of
`CO_2`

Now we have to calculate the mass of
`CO_2`

`\text{ Mass of }CO_2=\text{ Moles of }CO_2* \text{ Molar mass of }CO_2`

Molar mass of
`CO_2` = 44 g/mole

`\text{ Mass of }CO_2=(0.0208moles)* (44g/mole)=0.915g`

Therefore, the mass of
`CO_2` needed is, 0.915 grams.