Chem Class- January 26, 2010

Stoichiometry again. Last class we worked from mass to moles and today we worked from mass to mass.

Here's an example that we did in class: Lead (IV) nitrate reactions with 5.0 g of Potassium iodide. How many grams of Lead (IV) Nitrate are required?
Pb(NO3) + 4 KI -----> 4 KO3 + PbI4
5.0 g x 1 mol KI/165.9 g x 1 Pb(NO3)4/4KI x 455.2/ 1 mol Pb(NO3)4 = 3.4 g

All it is is one extra step, and that's converting the number of moles into grams. Here's some examples we put together:

Ex: C15H12 + 8 O2 -----> 5 CO2 + 6 H2O (Reaction when pentane burns). What mass of CO2 is produced when 100.0 g of C5H12 is burned?
100.0 g C5H12 x 1 mol C5H12/72.0 g C5H12 x 5 mol CO2/1 mol C5H12 x 44.0 g CO2/1 mol CO2 = 306 g

Ex: Given the reaction 4 CH3NO2 + 3 O2 -----> 4 CO2 + 6 H2O + 2N2, what mass of H2O is produced when 0.150 g of CH3NO2 is burned?
0.150 g CH3NO2 x 1 mol CH3NO2/61.0 g CH3NO2 x 6 mol H2O/4 mol CH3NO2 x 18.0 g H2O/ 1mol H2O = 0.0662 g

Ex:4NH3 + 5O2 -----> 6 H2O + 4 NO. Given this reaction, what mass of NO is produced when 2.00 mol of NH3 (g) are reacted with excess O2?
2.00 mol NH3 x 4 mol NO/ 4 mol NH3 x 30.0 g NO/ 1 mol NO = 60.0 g

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We also learned about:

PERCENT YIELD

The theoretical yield of a reaction is the expected (calculated) amount

The experimental amount is the actual yield

% yield = actual/theoretical x 100

Here's the example we did in class which required knowing the above information:

Example: The production of urea CO(NH2) is given by: 2NH3 + Co2 -----> CO(NH2) + H2O. If 47.7 g of urea are produced when 1 mole of CO2 reacts, find the actual yield, theoretical yield and percent yield.

We know that the actual yield is 46.7 g. The theoretical yield: 1 mol CO2 x 1 CO(NH2)2/1 CO2 x 60.1 g/ 1 mol = 60.1 g

Percent Yield = 47.7 g/60.1 g x 100 = 79.4%

Here's a video on calculating the percent yield: