Question

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

Answer 1

Answer:

The answer is "$SO_3 = 1.85 \times 10^{24} \ \ molecules$".

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,

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