Question

What mass of carbon dioxide (co2) can be produced from 15.6 g of c6h14 and excess oxygen?

Answer 1

C6H14+9.5O2=6CO2 +7H20
Number of moles of C6H14=15.6/86=0.1814 moles
so moles of CO2 = 6(0.1814)=1.088
As the c6h14 has 1 is to 6 ratio with co2
so
0.1814=mass/44
mass of co2 produced = 47.9 g

Answer 2

Answer: The mass of carbon dioxide produced is 46.7 grams.

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)

Given mass of hexane = 15.6

Molar mass of hexane = 88.2 g/mol

Putting values in equation 1, we get:

`\text{Moles of hexane}=(15.6g)/(88.2g/mol)=0.177mol`

The chemical equation for the combustion of hexane follows:

`2C_6H_(14)+19O_2\rightarrow 12CO_2+14H_2O`

As, oxygen gas is present in excess, it is considered as an excess reagent.

Thus, hexane is considered as a limiting reagent because it limits the formation of product

By Stoichiometry of the reaction:

2 moles of hexane produces 12 moles of carbon dioxide

So, 0.177 moles of hexane will produce =
`(12)/(2)* 0.177=1.062mol` of carbon dioxide

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

Molar mass of carbon dioxide = 44 g/mol

Moles of carbon dioxide = 1.062 moles

Putting values in equation 1, we get:

`1.062mol=\frac{\text{Mass of carbon dioxide}}{44g/mol}\\\\\text{Mass of carbon dioxide}=(1.062mol* 44g/mol)=46.7g`

Hence, the mass of carbon dioxide produced is 46.7 grams.