General Chemistry 10th Edition

Published by Cengage Learning
ISBN 10: 1-28505-137-8
ISBN 13: 978-1-28505-137-6

Chapter 3 - Calculations with Chemical Formulas and Equations - Exercises - Page 122: 3.90

Answer

0.59 g NO.

Work Step by Step

The strategy to find the mass of NO that is formed when we know the amount of the other product, 5.58 g Cu$(NO_{3})_2$ : g Cu$(NO_{3})_2$ ${\rightarrow}$ mol Cu$(NO_{3})_2$ ${\rightarrow}$ mol NO ${\rightarrow}$ g NO The conversion factors are: $\frac{1 mol (Cu(NO_{3})_2)}{Molar mass (Cu(NO_{3})_2)}$ to convert mass to moles Cu$(NO_{3})_2$, where Molar mass Cu$(NO_{3})_2$: Molar mass Cu$(NO_{3})_2$ = 1 x 64.0 + 2 x 14.0 + 6 x 16.0 = 188.0 g $\frac{2 mol (NO)}{3 mol (Cu(NO_{3})_2)}$, to convert the mol of Cu$(NO_{3})_2$ to mol of NO. $\frac{Molar mass(NO)}{1 mol(NO)}$ to convert to g of NO, where the molar mass of NO is: Molar mass NO = 1x 14 +1 x 16.0 = 30.0 g Now we use the conversion factors to find the mas of NO that is obtained, when we know the mass of 5.58 g of Cu$(NO_{3})_2$: 5.58 g Cu$(NO_{3})_2$ x $\frac{1 mol (Cu(NO_{3})_2)}{Molar mass (Cu(NO_{3})_2)}$ x $\frac{2 mol (NO)}{3 mol (Cu(NO_{3})_2)}$ x $\frac{Molar mass(NO)}{1 mol(NO)}$ = 5.58 g Cu$(NO_{3})_2$ x $\frac{1 mol (Cu(NO_{3})_2)}{188.0g (Cu(NO_{3})_2)}$ x $\frac{2 mol (NO)}{3 mol (Cu(NO_{3})_2)}$ x $\frac{30.0g (NO)}{1 mol(NO)}$ = 0.59 g NO
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