Question

How many grams of KClO3 are needed to produce 6.75 Liters of O2 gas measured at 1.3 atm pressure and 298 K?

Answer 1

Answer

29.3171 g KClO₃

Procedure

To solve this question consider the following chemical reaction:

2KClO₃→2KCl+3O₂

Then we will use the ideal gas formula to determine the number of moles present in 6.75 L of oxygen gas.

Data:

V=6.75 L

T= 298 °K

P= 1.3 atm

R= 0.08206 L⋅atm⋅°K⁻¹⋅mol⁻¹

Equation

PV=nRT

Solve for n to get the moles

$n=\frac{PV}{RT}=\frac{1.3\text{ atm }6.75\text{ L mol }\degree\text{K}}{0.08206\text{ L.atm 298}\degree\text{K}}=0.3588\text{ mol O}_2$

Use the stoichiometry of the reaction to convert from moles of oxygen to moles of KClO₃.

$0.3588\text{ mol O}_2\frac{2\text{ mol KClO}_3}{3\text{ mol O}_2}=0.2392\text{ mol KClO}_3$

Finally, convert from moles to grams

$0.2392\text{ mol KClO}_3\frac{122.55\text{ g KClO}_3\text{ }}{1\text{ mol KClO}_3}=\text{ 29.3171 g KClO}_3$