Chem Lecture 3 – Limiting Reactants, Percent Yield, Molar Mass, Percent Mass, Molarity – Flashcards
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molar mass
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the mass of one mole of a substance -read off the periodic table
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Molar mass of molecules
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sum of all the atomic masses
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mass percent
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mass of component/total mass
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empirical formula
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simplest formula, lowest whole number of moles --relative number of atoms Hydrogen peroxide H202(molecular formula)-> H0 (empirical formula)
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combustion
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burning of a substance with oxygen all carbons go to CO2 all hydrogens go to water
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stoichiometry
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relationship between quantities of reactants and products
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2Mg + O2 -> 2MgO How many moles MgO can be produced from 1.5moles of O2?
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(1.5molesO2 x 2molesMgO) / 1 mole O2 = 4=3 moles MgO
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2Mg + O2 -> 2MgO How many grams of Mg are needed to produce 2.25 moles MgO
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(2.25molesMgOx 2mole Mg)/ 2molesMgO) x 24.305gMg = 54.7 g Mg
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2Mg + O2 -> 2MgO How many moles of Mg are needed to completely react with 1.5gO2
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(1.5gO2x1moleO2)/32.00gO2 x (2molMg/1molO2) = 0.094mol Mg
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Limiting reactants
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the one that would produce least product 1. find moles of reactants (if in grams go to moles) 2. use mole ration to find moles of product each could make
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Excess (leftover) reactant
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Original amount - amount used -use mole ration how much leftover reactant was used
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theoretical yield
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max amount of product the limiting reactant can produce
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molarity, M
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concentraton unit moles solute/liters of solution
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moles
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MxV (in liters) molarityxvolume
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percent yield
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actual yield/theoretical yield
