Stoichiometry (Chapter 6) – Flashcards
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Avagodro's number =
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6.02 x 1023 (atoms, molecules, formula units)
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When converting atoms to Moles you must use
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Avagodro's number
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How many moles of HCl are in present in 20 atoms of HCl?
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Answer = 3.32 x 10-23 moles of HCl
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Molecular Weight
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Sum of atomic weights in a molecular compound
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What is the molecular weight of H20?
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18.02 amu
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Formula Weight
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Sum of atomic weights in an ionic compound
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What is the formula weight of NaCl?
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58.44 amu
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Molar Mass =
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Molecular weight or Formula weight
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If there are 206 g of MgCl2, how many moles are there?
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Answer = 2.16 moles of MgCl2
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When converting Molar Mass to Formula Units
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Molecular weight or Formula weight and avogadros number
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How many formula units of C9H8O4 (Aspirin) are in present in 45.0 g of aspirin?
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1.50 x 1023 formula units of C9H8O 4
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What is the step by step process to solving the following problem: How many formula units of C9H8O4 (Aspirin) are in present in 45.0 g of aspirin?
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45.0 g C9H8O4 ? Moles of C9H8O4 ?Formula Units of C9H8O4
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When converting Moles of one Atom to Moles of another Atom you must use the
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mole to mole ratio
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2CO + 1O2 ? 2CO2 How many moles of CO2 are in produced when 5.6 moles of O2 completely react?
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11.2 moles of CO2
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What is the step by step process to solving the following problem: How many grams of O2 are required to react with 64.9 grams of CO?
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64.9 g CO ?mol CO ? mol O2 ? g O2
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How many grams of O2 are required to react with 64.9 grams of CO?
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Answer = 37.07 g of O2
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2CO + 1O2 ? 2CO2 What is the step by step process to solving the following problem: What is the percent yield for the reaction above if 4.3 grams of CO2 is isolated in a reaction involving 64.9 grams of CO and excess O2?
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Step 1: Convert 64.9 grams of CO to grams of CO2 to find the theoretical yield Step 1: Plug in actual yield and theoretical yield into equation
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Percent yield = X 100
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Answer = 101.97 g of CO2
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Answer = 101.97 g of CO2
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(actual yield) / (theoretical yield) X 100
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What is the percent yield for the reaction above if 4.3 grams of CO2 is isolated in a reaction involving 64.9 grams of CO and excess O2?
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Answer = 4%
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Limiting Reagent=
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a reactant that runs out first in the reaction
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If 45 g of CO is reacted with 9 g of O2 what is the limiting reagent?
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Step 1: Convert 45 g of CO ? mol of CO Step 2: Convert 9 g of O2 ? mol of O2
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Answer = .28 mol O2
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Answer = .28 mol O2
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If there are 1.61 moles of CO .28 moles of O2 what is the limiting reagent?
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O2 is the limiting reagent
