Question

Combustion of amyl alcohol (C5H11OH) occurs according to the following

reaction :

2C5H11OH(l)+15O2(g) ® 10CO2(g)+12H2O(l)

(a) How many water molecules are produced from each mole of amyl alcohol

burnt?

(b) How much volume (in litre) of CO2 at STP (0 ºC, 1 bar pressure) is

produced from each gram of amyl alcohol burnt?

[Atomic mass : H = 1·0 u, C = 12·0 u, O = 16·0 u; N A = 6·022×10

23 m 1

ol

-

]​

Answer 1

Answer:

a) one mole of C5H11OH, will give 6 moles of water molecules

b) V = 0.131 L

Explanation:

2C5H11OH(l)+15O2(g) ® 10CO2(g)+12H2O(l)

Considering the above equation, we see that 2 moles of C5H11OH gives us 12 moles of water molecules. So when we have one mole of C5H11OH, then we will get 6 moles of water molecules.

First we calculate molar mass of C5H11OH = 12x5 + 1x11 + 16 + 1

Molar mass of C5H11OH = 88 g

Now if we have 1 grm of C5H11OH = 1/88 = 0.01136 moles will be available

Now again from the chemical equation, 2 moles of C5H11OH will give 10 moles of CO2, means 1 mole of C5H11OH will give 5 moles of CO2.

IF we have 1 grm (0.01136 moles) of C5H11OH, then moles of CO2 produced = 5* 0.01136 =0.0568

Using the ideal gas equation;

PV = nRT

P = 1 bar, T = 273K, R = 8.314 J / mol·K

V = 0.058 * 8.3148 * 273 = 131.64 m3

V = 0.131 L

Answer 2

Answer :

(a) The number of water molecules produced are $3.6\times 10^{24}$

(b) The volume of $CO_2$ at STP is 1.28 L

Explanation :

Part A :

The balanced chemical reaction is,

$2C_5H_{11}OH(l)+15O_2(g)\rightarrow 10CO_2(g)+12H_2O(l)$

From the balanced chemical reaction we conclude that,

As, 2 moles of amyl alcohol molecule react to give $12\times 6.022\times 10^{23}$ molecules of water

So, 1 moles of amyl alcohol molecule react to give $\frac{12\times 6.022\times 10^{23}}{2}=3.6\times 10^{24}$ molecules of water

Thus, the number of water molecules produced are $3.6\times 10^{24}$

Part B :

First we have to calculate the mass of $CO_2$.

From the balanced chemical reaction we conclude that,

As, $2\times 88g$ of amyl alcohol react to give $10\times 44g$ of $CO_2$

So, 1 g of amyl alcohol react to give $\frac{10\times 44g}{2\times 88g}=2.5g$ of $CO_2$

Now we have to calculate the moles of $CO_2$.

$\text{Moles of }CO_2=\frac{\text{Mass of }CO_2}{\text{Molar mass of }CO_2}=\frac{2.5g}{44g/mol}=0.057mole$

Now we have to calculate the volume of $CO_2$ at STP.

Using ideal gas equation,

$PV=nRT$

where,

P = pressure of gas = 1 bar = 1 atm

V = volume of gas = ?

T = temperature of gas = $0^oC=273+0=273K$

n = number of moles of gas = 0.057 mol

R = gas constant = 0.0821 L.atm/mol.K

Now put all the given values in the ideal gas equation, we get:

$(1atm)\times V=(0.057mol)\times (0.0821L.atm/mol.K)\times (273K)$

$V=1.28L$

Therefore, the volume of $CO_2$ at STP is 1.28 L