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 107: 16

Answer

There are 882. g of $NaHCO_3$

Work Step by Step

1. Find the number of moles of $NaHCO_3$: Each $NaHCO_3$ has 1 carbon. So: $10.5mol(C) \times \frac{1mol(NaHCO_3)}{1mol(C)} = 10.5 mol(NaHCO_3)$ 2. Determine the molar mass of this compound $(NaHCO_3)$: 22.99* 1 + 1.008* 1 + 12.01* 1 + 16.00* 3 = 84.01g/mol 3. Calculate the mass ($NaHCO_3$) $mm(g/mol) = \frac{mass(g)}{n(mol)}$ $mm(g/mol) * n(mol) = mass(g)$ $ 84.01 * 10.5 = mass(g)$ $882. = mass(g)$
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