Stoichiometry with Molarity and STP

Today we learned how to solve stoichiometry problems involving molarity and STP.


Remember Molarity =M


M=moles/Litres or moles= M/L  or moles/M = L

Example:
A tablet of Tums has a mass of 0.750 g. What volume of stomach acid(HCl) = 0.010 M is neutralized by a 0.750g portion of CaCO3?

1CaCO3(s) + 2 HCl(aq) ----> CaCl2(aq) + CO2(g) + H2O(l)

First find moles of HCl using stoichiometry.
0.750 g CaCO3 x 1 mole/100.1 g CaCO3 x 2 mol HCl/1 mol CaCO3 = 0.0150 mol

Now find the Litres using the molarity equation.
L= moles/Molarity
0.0150 mol/0.0010 mol/L = 15 L.

Now remember STP.

STP stands for Standard Temperature and Pressure

The conversion factor is 22.4 L / 1 mole or 1 mole/22.4 L.

You do the exact same thing with the other problems but when you have your desired substance in moles, you just multiply by 22.4 L / 1mole to find volume at standard temperature and pressure.

Example:
What volume of CO2(g) at STP is produced if 1.25 L of 0.0055 M HCl reacts with an excess of CaCO3?

First you find moles of the HCl.

moles = Molarity x litres

0.0055 M x 1.25 = 0.006875 mol HCl

Now just find the moles of CO2 using stoichiometry.

0.006875 x 1 mol CO2/ 2mol HCl

Now multiply by 22.4 L/ 1 mole.

Your answer will be 0.077 L CO2.

http://www.youtube.com/watch?v=z4nPxk1deuw

Here is some extra help!