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)

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 $O_2$.

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

$\text{Moles of }CH_4=\frac{8.00g}{16g/mol}=0.5mol$

and,

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

$\text{Moles of }O_2=\frac{16.00g}{32g/mol}=0.5mol$

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 $\frac{0.5}{2}=0.25$ 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 $\frac{0.5}{2}=0.25$ 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.

$\text{ Mass of }CH_4=\text{ Moles of }CH_4\times \text{ Molar mass of }CH_4$

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

and,

$\text{ Mass of }CO_2=\text{ Moles of }CO_2\times \text{ Molar mass of }CO_2$

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

and,

$\text{ Mass of }H_2O=\text{ Moles of }H_2O\times \text{ Molar mass of }H_2O$

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

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

Answer:

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

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