Question

The chemical equation below shows the combustion of methane (CH4).

CH4+2O2--CO2+2H2O

The molar mass of oxygen gas (O2) is 32.00 g/mol. The molar mass of carbon dioxide (CO2) is 44.01 g/mol. What mass of CO2, in grams, will firm when 8.94 g O2 completely react?
a)3.25
b)6.15
c)13.0
d)53.0

Answer 1

CH₄ + 2O₂ ---> CO₂ + 2H₂O
.............64g.........44g...............

64g O₂ --- 44g CO₂
8,94g O₂ --- X
X = (44×8,94)/64
X = 6,15g CO₂

b)

Answer 2

Answer: The correct answer is Option b.

Step-by-step explanation:

To calculate the number of moles, we use the equation:

`text{Number of moles}=\frac{text{Given mass}}{\text{Molar mass}}` .....(1)

• For Oxygen:

Given mass of oxygen = 8.94 g

Molar mass of oxygen = 32 g/mol

Putting values in above equation, we get:

`\text{Moles of oxygen}=(8.94g)/(32g/mol)=0.28mol`

The chemical equation for the combustion of methane is given as:

`CH_4+2O_2\rightarrow CO_2+2H_2O`

By Stoichiometry of the reaction:

2 moles of oxygen gas produces 1 mole of carbon dioxide.

So, 0.28 moles of oxygen gas will produce =
`(1)/(2)* 0.28=0.14mol` of carbon dioxide gas.

Now, calculating the mass of carbon dioxide by using equation 1.

Moles of carbon dioxide = 0.14 moles

Molar mass of carbon dioxide = 44.01 g/mol

Putting values in equation 1, we get:

`0.14mol=\frac{\text{Mass of carbon dioxide}}{44.01g/mol}\\\\\text{Mass of carbon dioxide}=6.15g`

Hence, the correct answer is Option b.