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CHAPTER 11
CHEMISTRY
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STOICHIOMETRYReview stoichiometry
CH4(g) + 2O2(g) --> 2H2O(g) + CO2(g)
If you had 5.0 moles of oxygen how many moles would you get of carbon dioxide?
If you had 85 g of methane how many moles of water would you form?
For every 5.0 grams of water how many grams of carbon dioxide would you have?
If you have 45 g of oxygen how many molecules of carbon dioxide would you form?
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LIMITING REACTANTSWhen one reactant has been completely used to form the product, and you still have some of the other reactant left over.2Na(s) + 2H2O(l) ---> 2NaOH(aq) + H2(g)If 90.0 g of sodium is dropped into 80.0 g water, how many grams of H2 would be produced? What is the limiting reactant? How much of the other reactant was left over?90.0gNa /2.0 g H2
/ 46.0 g Na3.90 g H2
80.0 gH2O / 2.0 g H2
/ 36.0 g H2O4.44 g H2
Sodium has to be the limiting reactant90.0gNa / 36.0 g H2O = / 46.0g Na70.4 so subtract 80.0 - 70.4 = 9.6 g H2O left over
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PERCENT YIELDIf you did an experiment where you start with 2.45 g Fe.Fe(s) + CuSO4(aq) --> FeSO4(aq) + Cu(s)How much copper did you form if you got a 73.4% yield? Actual yield X 100 = % yieldTheoretical yield2.45 g Fe / 63.5 g Cu = / 55.8 g Fe 2.79 g Cu X X 100 = 73.4% 2.79 g Cu X = 73.4/100 X 2.79 X = 2.05 g Cu
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MASS TO ENERGYCompare the heat of reaction for the displacement of 0.0663 g of sodium chloride from sodium bromide by chlorineCl2(g) + 2 NaBr(aq) --> 2NaCl(aq) + Br2(l) ΔH = - 100.18 kJ
Part of the product so you can use it in as a coefficient
0.0663g NaCl / 1 mol NaCl / -100.18 kJ =
/ 58.5 g NaCl / 2 mol
-0.0568 kJ