Chemistry 9th Edition

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

Chapter 3 - Stoichiometry - Exercises: 59

Answer

a. $3.54*10^{22}\text{ molecules $NH_3$}$ b. $1.88*10^{22}\text{ molecules $N_2H_4$}$ c. $2.39*10^{21}\text{ molecules $(NH_4)_2Cr_2O_7$}$

Work Step by Step

a. $1.00\text{ grams $NH_3$}*\frac{1\text{ mol $NH_3$}}{17.031\text{ grams $NH_3$}}*\frac{6.0221*10^{23}\text{ molecules $NH_3$}}{1\text{ mol $NH_3$}}=3.54*10^{22}\text{ molecules $NH_3$}$ b. $1.00\text{ grams $N_2H_4$}*\frac{1\text{ mol $N_2H_4$}}{32.0452\text{ grams $N_2H_4$}}*\frac{6.0221*10^{23}\text{ molecules $N_2H_4$}}{1\text{ mol $N_2H_4$}}=1.88*10^{22}\text{ molecules $N_2H_4$}$ c. $1.00\text{ grams $(NH_4)_2Cr_2O_7$}*\frac{1\text{ mol $(NH_4)_2Cr_2O_7$}}{252.0649\text{ grams $(NH_4)_2Cr_2O_7$}}*\frac{6.0221*10^{23}\text{ molecules $(NH_4)_2Cr_2O_7$}}{1\text{ mol $(NH_4)_2Cr_2O_7$}}=2.39*10^{21}\text{ molecules $(NH_4)_2Cr_2O_7$}$
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