Question

4. Suppose 8.00 g of CH4 is allowed to burn in the presence of 16.00 g of oxygen. CH4(g)+2O2(g)-->CO2(g)+2H2O(g)

How much (in grams) CH4, 02, CO2, and H2O (in grams) remain after the reaction is complete? ​

Answer 1

Answer: The mass of
`CH_4`,
`CO_2` and
`H_2O` remain after the reaction complete is, 4.0 g, 11 g and 9.0 g respectively.

Explanation : Given,

Mass of
`CH_4` = 8.00 g

Mass of
`O_2` = 16.00 g

Molar mass of
`CH_4` = 16 g/mol

Molar mass of
`O_2` = 32 g/mol

First we have to calculate the moles of
`CH_4` and
`O_2`.

`\text{Moles of }CH_4=\frac{\text{Given mass }CH_4}{\text{Molar mass }CH_4}`

`\text{Moles of }CH_4=(8.00g)/(16g/mol)=0.5mol`

and,

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

`\text{Moles of }O_2=(16.00g)/(32g/mol)=0.5mol`

Now we have to calculate the limiting and excess reagent.

The balanced chemical equation is:

`CH_4(g)+2O_2(g)\rightarrow CO_2(g)+2H_2O(g)`

From the balanced reaction we conclude that

As, 2 mole of
`O_2` react with 1 mole of
`CH_4`

So, 0.5 moles of
`O_2` react with
`(0.5)/(2)=0.25` moles of
`CH_4`

From this we conclude that,
`CH_4` is an excess reagent because the given moles are greater than the required moles and
`O_2` is a limiting reagent and it limits the formation of product.

The excess moles of
`CH_4` = 0.5 - 0.25 = 0.25 mol

Now we have to calculate the moles of
`CO_2` and
`H_2O`

From the reaction, we conclude that

As, 2 mole of
`O_2` react to give 1 mole of
`CO_2`

So, 0.5 moles of
`O_2` react with
`(0.5)/(2)=0.25` moles of
`CO_2`

and,

As, 2 mole of
`O_2` react to give 2 mole of
`H_2O`

So, 0.5 moles of
`O_2` react with 0.5 moles of
`H_2O`

Now we have to calculate the mass of
`CH_4`,
`CO_2` and
`H_2O` remain after the reaction is complete.

`\text{ Mass of }CH_4=\text{ Moles of }CH_4* \text{ Molar mass of }CH_4`

`\text{ Mass of }CH_4=(0.25moles)* (16g/mole)=4.0g`

and,

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

`\text{ Mass of }CO_2=(0.25moles)* (44g/mole)=11g`

and,

`\text{ Mass of }H_2O=\text{ Moles of }H_2O* \text{ Molar mass of }H_2O`

`\text{ Mass of }H_2O=(0.5moles)* (18g/mole)=9.0g`

Thus, the mass of
`CH_4`,
`CO_2` and
`H_2O` remain after the reaction complete is, 4.0 g, 11 g and 9.0 g respectively.

Answer 2

No masses of CH₄ and O₂ remained after the reaction, while 22.005 g of CO₂ and 18.02 g of H₂O remained

Step-by-step explanation:

The combustion of methane is given by the reaction;

CH₄(g)+2O₂(g) → CO₂(g)+2H₂O(g)

We are given, 8 g of CH₄ and 16.00 g of O₂

Required to determine the mass of CH₄, O₂, CO₂ and H₂O that remained after the complete reaction.

Step 1: Moles of CH₄ and O₂ in the mass given

Moles = mass ÷ molar mass

Molar mass of CH₄ = 16.04 g/mol

Moles of CH₄ = 8.00 g ÷ 16.04 g/mol

= 0.498 moles

= 0.5 moles

Molar mass of O₂ = 16.0 g/mol

Moles of O₂ = 16.00 g ÷ 16.00 g/mol

= 1 mole

From the reaction, 1 mole of CH₄ reacts with 2 moles of O₂

CH₄ is the limiting reactant since it is way less than the amount of O₂

Therefore, 0.5 moles of CH₄ will react with 1 mole of oxygen.

This means there will be no amount of O₂ and CH₄ that remains.

Step 2: Moles of CO₂ and H₂O that were produced.

From the reaction 1 mole of CH₄ reacts with 2 moles of O₂ to produce 1 mole of CO₂ and 2 moles of H₂O.

Therefore,

In our case, 0.5 moles of CH₄ will react with 1 mole of O₂ to produce 0.5 moles of CO₂ and 1 mole of H₂O.

Step 3: Mass of CO₂ and H₂O produced

Mass = Moles × Molar mass

Molar mass of CO₂ = 44.01 g/mol

Mass of CO₂ = 0.5 mol × 44.01 g/mol

= 22.005 g

Molar mass of H₂O = 18.02 g/mol

Moles of H₂O = 1 mole × 18.02 g/mol

= 18.02 g

Therefore, no masses of CH₄ and O₂ remained after the reaction, 22.005 g of CO₂ and 18.02 g of H₂O remained