Chemistry and Chemical Reactivity (9th Edition)

Published by Cengage Learning
ISBN 10: 1133949649
ISBN 13: 978-1-13394-964-0

Chapter 10 Gases and Their Properties - Study Questions - Page 403e: 67


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Work Step by Step

Number of moles of nickel: $0.450\ g\div 58.693\ g/mol=7.67×10^{-3}\ mol$ Number of moles of CO: $n=1.50\ L×(418/760×1\ atm)/0.082057\ L.atm/mol.K\div 298\ K$ $n=0.0337\ mol$ Ratio: $0.0337/0.00767=4.4\gt4$ CO is in excess Theoretical yield: $7.67×10^{-3}\ mol×170.7\ g/mol=1.31\ g$
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