Question

2. Consider the reaction 2 Cg H18 (4) +250â (9) ⺠16 co, (g) + 18 HâO(g) la How many moles of H20co) are produced, when |--16:15 moles of Ca Hiq react? answer: - mokes HâO | b. How many grams of oxygen are needed to completely react with 878 g of Cq Hig? answer: - g02

Answer 1

(a) The moles of water produced are 145.35 moles.

(b) The mass of oxygen needed are 3080.8 grams.

Solution for part (a) : Given,

Moles of
`C_8H_(18)` = 16.15 moles

First we have to calculate the moles of
`H_2O`

The balanced chemical reaction is,

`2C_8H_(18)+25O_2\rightarrow 16CO_2+18H_2O`

From the balanced reaction we conclude that

As, 2 moles of
`C_8H_(18)` react to give 18 moles of
`H_2O`

So, 16.15 moles of
`C_8H_(18)` react to give
`(16.15)/(2)* 18=145.35` moles of
`H_2O`

The moles of water produced are 145.35 moles.

Solution for part (b) : Given,

Mass of
`C_8H_(18)` = 878 g

Molar mass of
`C_8H_(18)` = 114 g/mole

Molar mass of
`O_2` = 32 g/mole

First we have to calculate the moles of
`C_8H_(18)`.

`\text{ Moles of }C_8H_(18)=\frac{\text{ Mass of }C_8H_(18)}{\text{ Molar mass of }C_8H_(18)}=(878g)/(114g/mole)=7.702moles`

Now we have to calculate the moles of
`O_2`

The balanced chemical reaction is,

`2C_8H_(18)+25O_2\rightarrow 16CO_2+18H_2O`

From the balanced reaction we conclude that

As, 2 moles of
`C_8H_(18)` react with 25 moles of
`O_2`

So, 7.702 moles of
`C_8H_(18)` react with
`(7.702)/(2)* 25=96.275` moles of
`O_2`

Now we have to calculate the mass of
`O_2`.

`\text{ Mass of }O_2=\text{ Moles of }O_2* \text{ Molar mass of }O_2`

`\text{ Mass of }O_2=(96.275moles)* (32g/mole)=3080.8g`

The mass of oxygen needed are 3080.8 grams.