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

Answer :

(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 $\frac{16.15}{2}\times 18=145.35$ 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 $C_8H_{18}$.

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

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 $\frac{7.702}{2}\times 25=96.275$ moles of $O_2$

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

$\text{ Mass of }O_2=\text{ Moles of }O_2\times \text{ Molar mass of }O_2$

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

The mass of oxygen needed are 3080.8 grams.