Answer
7.0 g of $N_{2}$
Work Step by Step
1 mol of $N_{2}$= 28.02 g of $N_{2}$ occupies 22.4 L volume at STP.
$\implies$ grams of $N_{2}$ in 5.6 L at STP=
$28.02\,g\times\frac{5.6\,L}{22.4\,L}=7.0\,g$
Chemistry: An Introduction to General, Organic, and Biological Chemistry (12th Edition)
by Timberlake, Karen C.
Published by
Prentice Hall
ISBN 10:
0321908449
ISBN 13:
978-0-32190-844-5
Chapter 8 - Section 8.6 - Volume and Moles (Avogadro's Law) - Study Check 8.8 - Page 271: 1
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