Question

Arctic acid (CH3COOH) is formed from its elements by the following reaction equation:

C(graphite) + 2 H2 + O2—-> CH3COOH
What is the volume of acetic acid CH3COOH gas produced by the reaction of 18.6 grams of H2 gas at 35 degrees Celsius and 1.05 atm?

Answer 1

1.05 atm is 106391.25 Pa

35 deg. C is 308 K

Number of moles of H2 gas involved in the reaction:

$n(H_2) = \frac{18.6 g}{2 \frac{g}{mol}} = 9.3 mol$

According to the reaction equation quantity in moles of produced acetic acid is two times less than H2. Thus,

$n(CH_3COOH) = \frac{9.3 mol}{2} = 4.65 mol$

Using general gas equation pV=nRT (where R is ideal gas constant) we are able to find the volume of acetic acid gas:

$V = \frac{4.65 mol \cdot 8.314 \frac{m^3 \cdot Pa}{K \cdot mol} \cdot 308 K}{106391.25 Pa} = 0.112 m^3 = 112 L$