Chemistry: An Introduction to General, Organic, and Biological Chemistry (12th Edition)

Published by Prentice Hall
ISBN 10: 0321908449
ISBN 13: 978-0-32190-844-5

Chapter 7 - Section 7.8 - Energy in Chemical Reactions - Additional Questions and Problems - Page 249: 7.97c

Answer

142 g of $F_2$ are needed to react with 25.5 g of $NH_3$

Work Step by Step

1. Calculate the molar mass of $NH_3$: $N: 14.01g $ $H: 1.008g * 3= 3.024g $ 14.01g + 3.024g = 17.03g $ \frac{1 mole (NH_3)}{ 17.03g (NH_3)}$ and $ \frac{ 17.03g (NH_3)}{1 mole (NH_3)}$ 2. The balanced reaction is: $2NH_3(g) + 5F_2(g) --\gt N_2F_4(g) + 6HF(g)$ According to the coefficients, the ratio of $NH_3$ to $F_2$ is 2 to 5: $ \frac{ 5 moles(F_2)}{ 2 moles (NH_3)}$ and $ \frac{ 2 moles (NH_3)}{ 5 moles(F_2)}$ 3. Calculate the molar mass for $F_2$: $F: 19.00g * 2= 38.00g $ $ \frac{1 mole (F_2)}{ 38.00g (F_2)}$ and $ \frac{ 38.00g (F_2)}{1 mole (F_2)}$ 4. Use the conversion factors to find the mass of $F_2$ $25.5g(NH_3) \times \frac{1 mole(NH_3)}{ 17.03g( NH_3)} \times \frac{ 5 moles(F_2)}{ 2 moles (NH_3)} \times \frac{ 38.00 g (F_2)}{ 1 mole (F_2)} = 142g (F_2)$
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.