Day 4 10-22 SOLVE: 1.5e24 molecules of SO 2 = ____ grams of SO 2 ? S 2 + 2O 2 2SO 2 And how many...
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Transcript of Day 4 10-22 SOLVE: 1.5e24 molecules of SO 2 = ____ grams of SO 2 ? S 2 + 2O 2 2SO 2 And how many...
Day 4 10-22
SOLVE: 1.5e24 molecules of SO2 = ____ grams of SO2?
S2 + 2O2 2SO2
And how many moles of S2?
Limiting Reactant – the reactant that limits the amounts of the other reactants that can combine and the amount of product that can form in a chemical reaction – the one you will run out of
Excess Reactant – any substance that is not used up completely in a reaction – left over
Limiting Reactants
Limiting Reactants
To determine the Limiting Reactant – use either reactant to calculate the amounts necessary for reaction
Example:
4NH3(g) + 6NO(g) 5N2(g) + 6H2O(l)
Given the following, determine the limiting reactant:
40 g NH3
41 g NO
Limiting ReactantsExample:
4NH3(g) + 6NO(g) 5N2(g) + 6H2O(l)
Given the following, determine the limiting reactant:
40 g NH3
41 g NO
2Na + Cl2 → 2NaCl
Limiting Reactants
Start with 4 moles of Na and 3 moles of Cl2:
limiting = ?
excess = ?
2Na + Cl2 → 2NaCl
Limiting Reactants
Start with 26 grams of Na and 35.5 grams of Cl2:
limiting = ?
excess = ?
Day 5 10-23
1. If you start with 2.5 moles of solid aluminum and allow it to react with 2.5 moles of CuCl2 which reactant will be left over at the end?
2. Why is today Mole Day?
Day 6 10-24
1. Which reactant determines your theoretical yield?
Percent Yield =
Percent Yield
Actual Yield Theoretical Yield
X 100
the percent yield is a measure of the efficiency of a reaction carried out in the lab. - a perfect percent yield would
be a 100%
Example # 1:Calcium carbonate, which is found in seashells, is decomposed by heating. The reaction is below:
CaCO3(s) CaO(s) + CO2(g)
What is the theoretical yield of CaO if 24.8 grams of CaCO3 is heated?
If only 12 grams of CaO are recovered, what is the percent yield?
What could be the cause?
Percent Yield = Actual Yield Theoretical Yield
X 100
12
Quarterly PracticeA solution containing 3.50 g sodium
phosphate is mixed with a solution containing 6.40 g barium nitrate. After the reaction is complete, 4.85 g barium phosphate are collected.
1. What is the limiting reagent?2. What is the theoretical yield of barium
phosphate?3. What is the percent yield of barium
phosphate?
2Na3PO4(aq) + 3Ba(NO3)2(aq) 6NaNO3(aq) + Ba3(PO4)2(s)
molar masses:Na3PO4 = 163.9Ba(NO3)2 = 261.3Ba3(PO4)2 = 601.9
13
Quarterly PracticeA solution containing 3.50 g sodium
phosphate is mixed with a solution containing 6.40 g barium nitrate. After the reaction is complete, 4.85 g barium phosphate are collected.
1. What is the limiting reagent?2. What is the theoretical yield of barium
phosphate?3. What is the percent yield of barium
phosphate?
2Na3PO4(aq) + 3Ba(NO3)2(aq) 6NaNO3(aq) + Ba3(PO4)2(s)
If you start with 3.50 g Na3PO4 you NEED 8.37 g Ba(NO3)2
So Barium nitrate is limiting and you start your theoretical with 6.40 g barium nitrateTheoretical yield = 4.91% yield = 98.8 %
Review sections 12.1 and 12.2 and complete # 8 on page 389, # 11 on page 391, # 13 on page 393, and # 23 on page 398 – due tomorrow
Assignment
Get to # 11
Hand in assignment from Friday
Review sections 12.1 and 12.2 and complete #s 5, 6, 8, 9, and 10 on page 389, #s 11 and 12 on page 391, and #s 13 and 14 on page 393 – due beginning of class Tuesday 10-29
Complete #s 21-25 on page 398 – due Wednesday 10-30
Assignment