Chemistry 10th Edition

Published by Brooks/Cole Publishing Co.
ISBN 10: 1133610668
ISBN 13: 978-1-13361-066-3

Chapter 3 - Chemical Equations and Reaction of Stoichiometry - Exercises - Calculations Based on Chemical Equations - Page 108: 25


682 g of $CoCl_2$ and 210 g $HF$ are necessary to produce 5.25 mol $CoF_2$.

Work Step by Step

- Cobalt (II) chloride $(CoCl_2)$: According to the balanced equation, the ratio of $CoF_2$ to $CoCl_2$ is 1 to 1: $5.25mol(CoF_2) \times \frac{1mol(CoCl_2)}{1mol(CoF_2)} = 5.25mol(CoCl_2)$ Molar mass $(CoCl2)$ = $58.93*1 + 35.45*2 = 129.83 g/mol$ - Calculate the mass of $CoCl_2$: $5.25mol(CoCl_2) \times \frac{129.83g(CoCl_2)}{1mol} = 682g(CoCl_2)$ ---------------------- - Hydrogen fluoride $(HF)$ According to the balanced equation the ratio of $CoF_2$ to $HF$ is 1 to 2: $5.25mol(CoF_2) \times \frac{2mol(HF)}{1mol(CoF_2)} = 10.5mol(HF)$ Molar mass $(HF): 1.008 + 19.00 = 20.01g/mol$ - Calculate the mass of $HF$: $10.5mol(HF) \times \frac{20.01g(HF)}{1mol(HF)} = 210g(HF)$
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