Chemistry 9th Edition

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

Chapter 5 - Gases - Exercises - Page 237: 72

Answer

1604.2 grams of $C_{3}$$H_{3}N$

Work Step by Step

PV=nRT P = .5MPa$\times$1000Kpa = 500Kpa T = 25 $c^{\circ}$ +273 = 298 Kelvin V = 150L R = 8.315 L*Kpa/K*mol (500Kpa)(150L)=n(8.315)(298) n = 30.26 mol $C_{3}$$H_{6}$ 30.26 mol $C_{3}$$H_{6}$ $\times$ $\frac{2 mol C_{3}H_{3}N }{2 mol C_{3}H_{6}}$ = 30.26 mol $C_{3}H_{3}N$ 30.26 mol $C_{3}H_{3}N$ $\times$ $\frac{53grams C_{3}H_{3}N}{1 mol}$ = 1604.2 grams $C_{3}H_{3}N$
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