1. By Stacy Rosete and Emily Guzman

2. How is stoichiometry used?  To convert from one unit to another unit, such as moles to moles, moles to grams, grams to grams, etc..

3. Grams to grams  How many grams of CuCl2 are needed to make NaCl if you have 25.6 g of CuCl2?  CuCl2 + NaNO3 → Cu(NO3)2 + NaCl  1) step up the table. (plug in the values and then numbers)  This will convert it to grams from the same compound to grams of the other one  ( 25.6 g of CuCl2)(moles of CuCl2/ g of CuCl2) (moles of NaCl/ moles of CuCl2) (g of NaCl/ moles of NaCl)

4. CuCl2 + NaNO3 → Cu(NO3)2 + NaCl  2) plug in the numbers in the equation  ( 25.6 g of CuCl2)(1 moles of CuCl2/ 134.45g of CuCl2) ( 1moles of NaCl/ 1moles of CuCl2) (58.45g of NaCl/ 1 moles of NaCl)  Note: we will be using the molar mass of both compounds. When its moles over moles we will be using the numbers in front of the equation, in this case one.  Solve! Answer is 11.13 g of NaCl

5. Gas stoichiometry  Hydrogen gas (and NaOH) is produced when sodium metal is added to water. What mass of Na is needed to produced 10.0 L of H2 at STP?  1) need a balance equation:  2Na(s) + 2H2O(l) → H2(g) + N2aOH(aq)  2) then use PV=nRT to solve for moles of H2  P= 101.3 kPa, V= 10.0 L, T= 273 K, R=8.3  3) plug this numbers in PV=nRT  101.3(10)= n(8.3)(273)  N=2.24 moles of H2

6. 2Na(s) + 2H2O(l) → H2(g) + N2aOH(aq)  4) convert moles to grams of Na  Apply the basic steps  (2.24 g of H2)(2 mol of Na/ 1 mol H2)(22.99/1 mol Na)  102.3 g of Na

7. More Gas stoichiometry  redo

8. 2Al(s)+ 6HCl(aq)  2AlCl3(aq) + 3H2(g)  How many grams of aluminum( atomic mass 27) are necessary to produce 4 mol of hydrogen gas at 25 C and 1.00 atm?  1) convert the 4 moles of hydrogen gas to g of Al  (4 mol of H2)(2mol of Al/3 mol H2)(27 g off Al)= 72 g of Al

9. Solution stochiometry  How many mL of 0.150 M NaCl solution contains 2.45 g of NaCl?  Set up grams to moles stoichiometry by placing the given 2.45 g of NaCl times 1 mole of NaCl over the total molar mass of NaCl that is 58.45g.

10.  Divide 2.45g by 58.45g to equal 0.0419 moles. Use the formula mol = VM, volume in Liters to solve. Use 0.0419 mol times 1L over 0.150 M of NaCl and divide.

11.  Since we are solving for mL, we multiply the result by 1000 to get 279 mL.

12. More Solution stoichiometry.. redo  Mix 300. ml of 0.100 M Nacl solution with excess Pb(NO3)2, how many grams of PbCl2 precipitate will be formed?  1) we will be converting 300 ml to liters =.300 L of 0.100 M NaCl  2) convert L of NaCl to g of PbCl2  (tip: how are liters and Molarity related? Well molarity is moles over liters, use the molarity to convert the solution from moles to grams)  (.300 l NaCl)(.100 mol of NaCl/1 L of NaCl)(1 mol of AgCl/1 mol of AgNO3)( 143.3212 g AgCl/1 mols of AgCl)= 5.32 g of AgCl