Chemistry: Atoms First (2nd Edition)

Published by Cengage Learning
ISBN 10: 1305079248
ISBN 13: 978-1-30507-924-3

Chapter 5 - Exercises - Page 240d: 57

Answer

a. 70.1 g of nitrogen b. 140. g of nitrogen c. 140. g of nitrogen

Work Step by Step

a. - Each $ NH_3 $ has 1 N atoms, thus:$$ 5.00 \space mole \space NH_4 \times \frac{ 1 \space moles \ N }{1 \space mole \space NH_3 } = 5.00 \space moles \space N $$ $ N $ : 14.01 g/mol - Using the molar mass as a conversion factor, find the mass in g: $$ 5.00 \space mole \times \frac{ 14.01 \space g}{1 \space mole} = 70.1 \space g$$ b. - Each $ N_2H_4 $ has 2 N atoms, thus:$$ 5.00 \space mole \space N_2H_4 \times \frac{ 2 \space moles \ N }{1 \space mole \space N_2H_4 } = 10.0 \space moles \space N $$ $$ 10.0 \space mole \times \frac{ 14.01 \space g}{1 \space mole} = 140. \space g$$ c.- Each $ N_2H_8Cr_2O_7 $ has 2 N atoms, thus:$$ 5.00 \space mole \space N_2H_8Cr_2O_7 \times \frac{ 2 \space moles \ N }{1 \space mole \space N_2H_8Cr_2O_7 } = 10.0 \space moles \space N $$ $$ 10.0 \space mole \times \frac{ 14.01 \space g}{1 \space mole} = 140. \space g$$
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