Question

Acetic acid (CH3COOH) is formed from its elements by the following reaction equation:
2 C (graphite) + 2 H2 (g) + O2 (g) → CH3COOH (g)1. What is the volume of acetic acid CH3COOH gas produced by the reaction of 18.6 grams of H2 gas at 35°C and 1.05 atm?

Answer 1

Answer:

111 L

Explanation:

Calculation of moles of hydrogen gas:-

Mass of $H_2$ = 18.6 g

Molar mass of $H_2$ = 2.01588 g/mol

$Moles=\frac{Mass}{Molar\ mass}=\frac{18.6}{2.01588}\ mol=9.23\ mol$

According to the given reaction:-

$2C+2H_2+O_2\rightarrow CH_3COOH$

2 moles of hydrogen gas on reaction produces one mole of acetic acid gas.

So,

1 mole of hydrogen gas on reaction produces $\frac{1}{2}$ mole of acetic acid gas.

Also,

9.23 mole of hydrogen gas on reaction produces $\frac{1}{2}\times 9.23$ mole of acetic acid gas.

Moles of acetic acid gas = 4.615 moles

Given that:

Temperature = 35 °C

The conversion of T( °C) to T(K) is shown below:

T(K) = T( °C) + 273.15  

So,  

T = (35 + 273.15) K = 308.15 K  

n = 4.615 moles

P = 1.05 atm

V = ?

Using ideal gas equation as:

$PV=nRT$

where,  

P is the pressure

V is the volume

n is the number of moles

T is the temperature  

R is Gas constant having value = 0.0821 L atm/ K mol  

Applying the equation as:

1.05 atm × V = 4.615 moles ×0.0821 L atm/ K mol  × 308.15 K  

⇒V = 111 L