Lecture_04OCT

# Lecture_04OCT - Calculations from chemical equations If you...

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Unformatted text preview: Calculations from chemical equations If you know the amount of any reactant or product involved in the reaction: • you can calculate the amounts of all the other reactants and products that are consumed or produced in the reaction C 3 H 8 ( g ) + 5 O 2 ( g ) 3 CO 2 ( g ) + 4 H 2 O( g ) BUT REMEMBER ! The coefFcients in a chemical equation provide information ONLY about the proportions of MOLES of reactants and products • given the number of moles of a reactant/product involved in a reaction, you CAN directly calculate the number of moles of other reactants and products consumed or produced in the reaction • given the mass of a reactant/product involved in a reaction, you can NOT directly calculate the mass of other reactants and products consumed or produced in the reaction Mass - mass calculations • A balanced chemical equation • A known mass of one of the reactants/product (in grams) Given: Calculate: • The mass of one of the other reactants/products (in grams) Use ratio between coefficients of substances A and B from balanced equation Moles of substance A Moles of substance B Grams of substance A Grams of substance B Use molar mass of substance A Use molar mass of substance B Mass - mass calculation: Another example How many grams of carbon dioxide are produced by the complete combustion of 100. g of pentane (C 5 H 12 ) ? The balanced equation is: C 5 H 12 ( g ) + 8 O 2 ( g ) 5 CO 2 ( g ) + 6 H 2 O( g ) 100. g C 5 H 12 ( 1 mol C 5 H 12 / 72.15 g C 5 H 12 ) = 1.39 mol C 5 H 12 Step 1: Convert the amount of known substance ( C 5 H 12 ) from grams to moles Molar mass C 5 H 12 : ( 5 x 12.01 g/mol ) + ( 12 x 1.008 g/mol ) = 72.15 g/mol Mole ratio between the unknown substance ( carbon dioxide ) and the known substance ( pentane ): 5 mol CO 2 1 mol C 5 H 12 5 mol CO 2 1 mol C 5 H 12 n 1.39 mol C 5 H 12 = ( 1.39 mol C 5 H 12 ) ( 1.39 mol C 5 H 12 ) n = 6.95 mol CO 2 Step 2: Determine the number of moles of the unknown substance ( CO 2 ) required to produce the number of moles of the known substance ( 1.39 mol C 5 H 12 ) How many grams of carbon dioxide are produced by the complete combustion of 100. g of pentane (C 5 H 12 ) ? The balanced equation is: C 5 H 12 ( g ) + 8 O 2 ( g ) 5 CO 2 ( g ) + 6 H 2 O( g ) Mass - mass calculation: Another example How many grams of carbon dioxide are produced by the complete combustion of 100. g of pentane (C 5 H 12 ) ? The balanced equation is: C 5 H 12 ( g ) + 8 O 2 ( g ) 5 CO 2 ( g ) + 6 H 2 O( g ) n = 6.95 mol CO 2 ( 44.01 g CO 2 / 1 mol CO 2 ) = 306 g CO 2 Step 3: Convert the amount of unknown substance ( 6.95 moles CO 2 ) from moles to grams Molar mass CO 2 : 12.01 g/mol + ( 2 x 16.00 g/mol ) = 44.01 g/mol Mass - mass calculation: Another example The concept of limiting reactants In some chemical reactions, all reagents are present in the exact amounts required to completely react with one another.amounts required to completely react with one another....
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Lecture_04OCT - Calculations from chemical equations If you...

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