Chemistry 9th Edition

Published by Cengage Learning
ISBN 10: 1133611095
ISBN 13: 978-1-13361-109-7

Chapter 4 - Types of Chemical Reactions and Solution Stoichiometry - Exercises - Page 183: 55


There are necessary 0.607 g of $Na_2CrO_4$ to precipitate all the silver ions.

Work Step by Step

- Write the reaction between $Na_2CrO_4$ and $AgNO_3$: $Na_2CrO_4(aq) + AgNO_3(aq) -- \gt Ag_2CrO_4(s) + NaNO_3(aq)$ - Balance the reaction. Begin with the number of sodium (Na) atoms: $Na_2CrO_4(aq) + AgNO_3(aq) -- \gt Ag_2CrO_4(s) + 2NaNO_3(aq)$ - Balance the number of $Ag$ atoms: $Na_2CrO_4(aq) + 2AgNO_3(aq) -- \gt Ag_2CrO_4(s) + 2NaNO_3(aq)$ - The equation is balanced. 1. The molar mass for $Na_2CrO_4$ is: 22.99* 2 + 52* 1 + 16.00* 4 = 161.98g/mol $(Na_2CrO_4)$ 2. According to the balanced equation, each mole of $Na_2CrO_4$ reacts with 2 moles of $AgNO_3$. 3. The molarity of the $AgNO_3$ is equal to $0.100mol/L$ 4. 1000 mL = 1 L --------------------------- Now, use these informations as conversion factors to find the $Na_2CrO_4$ required mass. $75.0mL \times \frac{1L}{1000mL} \times \frac{0.100mol(AgNO_3)}{1L} \times \frac{1mol(Na_2CrO_4)}{2mol(AgNO_3)} \times \frac{161.98g(Na_2CrO_4)}{1mol(Na_2CrO_4)} = 0.607 g (Na_2CrO_4)$
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