# Question 6f3a1

Mar 15, 2015

This time, your reaction will produce $\text{6530 mL}$ of oxygen gas.

I'll build this answer upon a previous one I gave to a similar problem

http://socratic.org/questions/decomposition-of-kclo3-serves-as-convenient-lab-source-of-small-amounts-of-o2-th132616

So, you've determined that 23.5 g of potassium chlorate will produce 0.2877 moles of oxygen gas.

This time, the conditions for temperature and pressure have changed. Standard Temperature and Pressure, or STP, is defined as having a pressure equal to 100 kPa and a temperature of $\text{0"^@"C}$.

The interesting part about these conditions is that, at STP, 1 mole of any ideal gas occupies exactly 22.7 L - this is known as the molar volume of a gas at STP.

Since you've got less than 1 mole of oxygen gas, you'll have less than 22.7 L.

$n = \frac{V}{V} _ \left(\text{molar") => V = n * V_("molar}\right)$

$V = \text{0.2877 moles" * "22.7 L" = "6.53 L}$

Expressed in mililiters, the answer will be

$\text{6.53 L" * "1000 mL"/"1 L" = "6530 mL}$

SIDE NOTE Notice that this time the volume is smaller compared with the volume obtained at room temperature and 1.25 atm.

The decrease in temperature would have made the volume smaller, but this effect was overpowered by the decrease in pressure $\to$ overall, the volume dropped because the drop in pressure had a bigger impact than the drop in temperature.