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1 CH 4 + H 2 O 3H 2 + CO If you have 32 g CH 4 and 32 g H 2 O, which - PDF document

Sep 24, 2022 •476 likes •524 views

Unbalanced equation sample + Reactant A Reactant B Product C Balanced equation = A + B C 2 sample 3 x C Excess Are any reactants leftover ? 1 x B reactant How many molecules of product can be formed Which reactant

  1. Unbalanced equation sample + → Reactant A Reactant B Product C Balanced equation = A + B → C 2 sample 3 x C Excess Are any reactants leftover ? 1 x B reactant How many molecules of product can be formed Which reactant determined the with the sample given? A amount of product formed? The reactant that determines the amount of product formed is called the limiting reactant sample + + + → Reactant A Reactant B Products C + D Balanced equation =A + B → C + D sample 4 x C, 4 x D Is anything leftover? Are any reactants limiting ? no Reactants A + B are in correct ratio as needed by balanced equation = the reactants are in a stoichiometric ratio CH 4 + H 2 O � 3H 2 + CO CH 4 + H 2 O � 3H 2 + CO 3 mole CH 4 and 2 moles H 2 O how many moles of CO can be produced? If you have 1 mole CH 4 and 2 moles H 2 O , which is limiting reactant? 2 mole CH 4 3 mole CH 4 and 2 moles H 2 O how many moles of H 2 If you have 1 mole CH 4 and 2 moles H 2 O how can be produced? many moles of CO can be produced? 6 mole 1 mole 1

  2. CH 4 + H 2 O � 3H 2 + CO If you have 32 g CH 4 and 32 g H 2 O, which is limiting? Two types of LR problems: Convert to moles to ID LR: 1. Solving for amount of product formed � � � 1 mole CH 4 32 g CH 4 � = 2 mole CH 4 � � 4 16 g CH 4 � � � � � � � O 1 mole H 2 O 32 g H 2 O � = 1.77 mole H 2 O � � � 18 g H 2 O � � � � 1 mole CH 4 Balanced eq. requires ratio 1 mole H 2 O H 2 O = LR Will run out of H 2 O before CH 4 N 2 + 3H 2 � 2NH 3 32.0 g of nitrogen gas are reacted with 5.92 g hydrogen gas to form ammonia. How much ammonia Step 3: Compare actual ratio (what you have) to forms? what equation requires (what you need) to Step 1: Balance equation determine LR: N 2 + 3H 3H 3H 2 � 2NH 2NH 2NH 3 you have: simplify 2.96 mole H 2 = 2.6 mole H 2 larger Step 2: Convert g to moles 1.14 mole N 2 1 mole N 2 smaller � � � 1 mole N 2 � 32 g N 2 � = 1.14 mole N 2 Compare � � � 2 28 g N 2 ratios � � � � You need: H 2 = LR 3 mole H 2 � � � � 1 mole H 2 5.92 g H 2 � = 2.96 mole H 2 1 mole N 2 � � � 2 2 g H 2 � � � � Step 4: Use # moles of LR to perform stoichiometry (LR determines amount of product formed) 2. Solving for amount of excess N 2 + 3H 2 � 2NH 3 H 2 = LR reactant remaining � � � � � � � 2 mole NH 3 � 17 g NH 3 2.96 mole H 2 � = 33.5 g NH 3 � � � � � 2 3 mole H 2 1 mole NH 3 � � � � � � � 2

  3. 26 g of oxygen are reacted with 3 g hydrogen to Step 3: Compare actual ratio you have to what form water. How much of the excess reactant equation requires to determine LR: remains after the reaction is complete? You have: Step 1: Balance equation 1.5 mole H 2 = 1.85 mole H 2 2H 2H 2 + O 2 � 2H 2H 2 O 2H 0.813 mole O 2 1 mole O 2 Step 2: Convert g to moles � � � You need: 1 mole O 2 26 g O 2 � = 0.813 mole O 2 � � 2H 2 + O 2 � 2H 2 O 2 32 g O 2 � � � 2 mole H 2 � � 1 mole H 2 � � 3 g H 2 � = 1.5 mole H 2 � � � H 2 = LR 2 1 mole O 2 2 g H 2 � � � � Step 4: Use # moles of LR to perform 26 g - 24 g = 2 g stoichiometry to determine how much of other reactant used in the reaction Starting how much O 2 mass O 2 mass O 2 used remaining 2H 2 + O 2 � 2H 2 O LR excess Calc. how much O 2 consumed: � � � � � � � 1 mole O 2 � 32 g O 2 1.5 mole H 2 � = 24 g O 2 � � � � � 2 2 mole H 2 1 mole O 2 � � � � � � � LR moles Used in rxn 3

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