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

111 L

Step-by-step explanation:

Calculation of moles of hydrogen gas:-

Mass of
`H_2` = 18.6 g

Molar mass of
`H_2` = 2.01588 g/mol

`Moles=(Mass)/(Molar\ mass)=(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
`(1)/(2)` mole of acetic acid gas.

Also,

9.23 mole of hydrogen gas on reaction produces
`(1)/(2)* 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