Transcript Stoichiometry

By Stacy Rosete and Emily Guzman
STOICHIOMETRY
REDO
How is stoichiometry used?
 To convert from one unit to another unit,
such as moles to moles , moles to grams,
grams to grams , etc..
Grams to grams
 How many grams of CuCl2 are needed to
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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)
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
Gas stoichiometry
 Hydrogen gas (and NaOH) is produced when
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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
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
More Gas stoichiometry
 redo
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
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.
 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.
 Since we are solving for mL, we multiply the
result by 1000 to get 279 mL.
More Solution
stoichiometry.. redo
 Mix 300. ml of 0.100 M Nacl solution with excess
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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