Stoichiometry with Molarity and STP

Here’s a review of the Molarity formula if you don’t remember:

Molarity=   Moles  .

                  Volume

which can also be written as:

Volume=  Moles  .         OR          Moles= Molarity X Volume

                Molarity

Volume must be measured in Litres.

Let’s try an example:

Consider this chemical equation: NiCl3(aq) + Ca(s) → CaCl2(aq) + Ni(s)

How many grams of nickel will be formed when 110mL of 0.270M of nickel (III) chloride is reacted with sufficient calcium?

STEP 1: balance the equation

2 NiCl3(aq) + 3 Ca(s) à 3 CaCl2(aq) + 2 Ni(s)

STEP 2: make a mole map (for visual purposes)

110mL of 0.270M NiCl3 -------------------------------------> ? grams Ni

                                 -----> moles NiCl3 -----> moles Ni ----->

STEP 3: calculate

110mL = 0.110L        ßconvert mL to L

mol = M x L

0.110L NiCl3 X 0.270M NiCl3  = 0.0297mol NiCl3    ßfind mole

0,0297mol NiCl3 X  2mol Ni   = 0.0297mol Ni   ßconvert using mole ratio

                               2mol NiCl3

0.0297mol Ni X 58.7g Ni  = 1.7g Ni   ßconvert to grams

                            1mol Ni

1.7g of nickel will be formed.

Here’s another example:

How many mL of 0.200M NiCl3 are required to react with 2.10 grams of calcium?

STEP 1: chemical equation is already balanced in previous example

STEP 2:  

2.10g Ca -----------------------------------> ? mL NiCl3

        ----> moles Ca ----> moles NiCl3 ----->

STEP 3:

2.10g Ca X 1mol Ca    X  2mol NiCl3  = 0.0349mol NiCl3

                    40.1g Ca          3mol Ca

V=.   mol   .        0.0349mol NiCl3  =  0.175L NiCl3

      Molarity          0.200M NiCl3

0.175L = 175mL

175mL of NiCl3 are required.

STP: Standard Temperature and Pressure

                         22.4L

                         1mol

            ------------------------->

MOLES                                STP

           <--------------------------

                         1mol

                         22.4L

Example:

Sodium carbonate decomposes into carbon dioxide and sodium oxide. How many grams of sodium carbonate are needed to produce 11.3L of carbon dioxide measured at STP?

STEP 1: find out and balance the equation

1 Na2CO3 à 1 CO2 + 1 Na2O

STEP 2: calculate

Convert the volume of CO2 into moles using STP formula

11.3L CO2 X 1mol CO2  =  0.504mol CO2

                      22.4L CO2

Use mole ratio

0.504mol CO2 X 1mol Na2CO3  =  0.504mol Na2CO3

                                1mol CO2

Convert moles to grams

0.504mol Na2CO3 X 106g Na2CO3  =  53.5g Na2CO3

                                  1mol Na2CO3

53.5 grams of Na2CO3 are needed.

Here are a few worksheets on stoichiometry and molarity:

http://misterguch.brinkster.net/PRA048.pdf