General Types of Reactions Notes

# How to do the conversion massg of moles x molar mass

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How to do the conversion: Mass(g) = (# of moles) x Molar Mass (g/mol) <--Same Mass(g)/ Molar Mass (g/mol)= Mole Moles x Avogadro's # atoms or molecules (6.022 x 1023)=# of atoms or molecules Grams <---> Moles <---> atoms or Formula units molecules (below is another rep. with math x and /) Grams atoms or molecules (molar mass)(divide) Moles (Avogadro's #) (multi)

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How many molecules of sugar are present in a teaspoon of sugar, C12H32C11, that weighs 10.5 g? MM=342.34 g/mol done by mult. Number of each and average weight 10.5g C 12 H 32 C 11 x (1 mole/342.34g) x (6.022x 10^23 molecules/1 mole) = 1.85 x 10 22 molecules of C 12 H 32 C 11 How many Carbon atoms are present? 1.85x 10 23 molecule C 12 H 32 C 11 x (12 C atoms/1 molecule C 12 H 32 C 11) = 2.22 x 10 23 atoms 2NaN3 --> 2Na +3N2 (2 and 3 are called stoichicmetric coe ffi cients) Stoichio means element and metric means measure Not based on mass and Is based on the numbers (# or mole) 4 mol NaN3 x 3 mol N2/ 2 mol NaN3 = 6 mol N2 (this section is called stoichiometric ratio mole:mole ratio) Mass A MassB (molar mass)(divide) MolesA----------(mole B / moleA)------------------ MolesB (molar mass) IN Fig 13 How many grams of H2O will be produced in combustion of 10.0g C3H8? C3H8 + 5O2 ---> 3CO2 + 4H2O 10.0g C3H8 x (1 mol C3H8/ 44.81 C3H8) x (4 mol H2O/1 mol C3H8) x (18.02 g H2O/ 1 mol H2O) = 16.3 g H2O If 0.0308 g CO2 were produced, how many grams of O2 were consumed? 0.0308g CO2 x (1 mole CO2/ 44.01 g CO2) x (5 mole O2/3mole CO2) x (32.00g O2/1 mole O2) = 0.0373g O2 Molar Mass --> mass of substance in 1 mole 32.00 g O2 = 1 mole O2 2H2 + O2 ---> 2 H2O Initial (10mole) + 7 mole---->0 mole 1o mole H2 x (1 mol O2/ 2 mole H2) = 5 mole O2 will be used 1o mole H2 x (2 mol H2O/ 2 mole H2) = 10 mol H2O will be produced Reactant (-) in cone Products in (+) concentration
Limiting Reagent --> use of stoich coefi to predict amts of reactants and products except amounts of 2 reactants are given. Work problem mass1 ---> Massproduct and mass2 ----> Massproduct Whichever Reactant (1 or 2) that produces the smaller amt of product is limiting Reagent. The smaller amount of product is theoretical yield . The larger amount of product is WRONG.

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