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114 Chapter Three Stoichiometry ./ 68.5 kg CO(g) ./ 8.60 kg Hig) ./ 3.57 X 10 4 g CH 3 0H is produced What information do we need? ./ Balanced equation for the reaction ./ Moles of H2 ./ Moles of CO ./ Which reactant is limiting ./ Amount of CHPH produced How do we get there? To find the limiting reactant What is the balanced equation? 2Hig) + CO(g) ~ CH 3 0H(l) What are the moles of H2 and CO? To find the moles, we need to know the molar masses. H2 2.016 glmol CO 28.02 glmol 1000 g.-eo X 1 mol CO = 2.44 X 10 3 mol CO 68.5 kg-eeJ X 1 kg-eeJ 28.02 g.-eo 1000 %-H2 X 1 mol H2 = 4.27 X 10 3 mol H2 8.60 kg-H2 X 1 kg-H2 2.016 %-H2 What is the mole ratio between H2 and CO in the balanced equation? 2 mol H2 I mol CO How does the actual mole ratio compare to the stoichiometric ratio? To determine which reactant is limiting, we compare the mole ratio of H2 and CO re- quired
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Unformatted text preview: by the balanced equation mol H2 2-m-o-lC-O'='" (required) = 1 = 2 with the actual mole ratio mol H2 4.27 X 10 3-m-o-lC-O'='" (actual) = 2.44 X 10 3 = 1.75 • Since the actual mole ratio of H2 to CO is smaller than the required ratio, H2 is limiting. To calculate the theoretical yield of methanol What are the moles of CH 3 0H formed? We therefore must use the amount of H2 and the mole ratio between H2 and CH 3 0H to determine the maximum amount of methanol that can be produced: 1 molCH 3 0H 4.27 X 10 3 meHI 2 X = 2.14 X 10 3 mol CH 3 0H 2meHI 2 What is the theoretical yield of CH 3 0H in grams? 32.04 g CH OH 2.14 X 10 3 ~ X 3 = 6.86 X 10 4 g CH 3 0H 1~...
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This note was uploaded on 12/13/2010 for the course CHEM 2301 taught by Professor Bill during the Spring '10 term at South Texas College.

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