Question

How many molecules of sulfur trioxide would react with 68.9 L of water at STP to produce sulfuric acid?

Answer 1

The answer is "
`SO_3 = 1.85 * 10^(24) \ \ molecules`".

Step-by-step explanation:

In the given question some information is missing, that chemical reaction, which can be described as follows:

`SO_3+H_2O= H_2SO_4`

where,

`SO_3 =` Sulfur trioxide

`H_2O =` water

`H_2SO_4`= Sulfuric acid

calculating molecules of sulfur trioxide:

Formula:

`PV =nRT \\`

Where,

`P= 1 \\\ T= 273.15 K\\R= 0.08206 \ \ (l)/(mol) \\V = 68.9 l\\n= ?\\\\n =(PV)/(RT) \\\\O_2= (1 * 68.9)/(0.08206 * 273.15) \\\\O_2 = 3.7 mole\\\\\ if \ mole \ ratio \ is \ 1: 1 \ where SO_3 : H_2O \\\\\therefore SO_3 = 3.07 mole \\\\1 \ mole \ of \ any \ subtance = 6. 022 * 10^(23) molecule \\\\\ for \ n \ molecule = 6. 022 * 10^(23) * 3.07 \\\\ SO_3 = 1.85 * 10^(24) molecules`