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 - Using Solutions in Chemical Reactions - Page 111: 80


8.41 mL of that 0.558 M $HNO_3$ solution.

Work Step by Step

1. The concentration of the $Ba(OH)_2$ solution is equal to 0.0515 M. 2. The concentration of the $HNO_3$ solution is equal to 0.558 M. 3. According to the balance coefficients, each mol of $Ba(OH)_2$ reacts with 2 moles of $HNO_3$. 4. 1000 mL = 1 L - Use this information as conversion factors to find the volume of $HNO_3$ solution we should use to react completely with 45.55 mL of that $Ba(OH)_2$ solution. $45.55mL (Ba(OH)_2) \times \frac{1L}{1000mL} \times \frac{0.0515mol(Ba(OH)_2)}{1L(Ba(OH)_2)} \times \frac{2 mol (HNO_3)}{1mol(Ba(OH)_2)} \times \frac{1L(HNO_3)}{0.558mol(HNO_3)} = 0.00841 L (HNO_3)$ $0.00841L \times \frac{1000mL}{1L} = 8.41mL$
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.