Consider the following reaction: 4K(s)+O2(g)→2K2O(s)4K(s)+O2(g)→2K2O(s)The molar mass of KK is 39.10 gmol−1gmol−1 and that of O2O2 is 32.00 gmol−1gmol−1.The limiting reactant is 1.5 g K, 0.38 g O2I need help with the "explain your reasoning" 5The molar mass of K isO2. Since the stoichiometry has a ration of4mol K tomol 02, K will be the limiting reactant when mass of K ismuch greater thanthantimes the mass of O2 .much lower thanfourfive1comparable to6lessthree2moretwosix

The reaction taking place is given as, => 4 K (s) + O2 (g) → 2 K2O (s)

The molar mass of K is O2. Since the stoichiometry has a ration of 4 mol K to mol O2 , K will be the limiting reactant when mass of K is much greater than than times the mass of O2 - 3 much lower than four five 1 comparable to 6 less three 2 more two six

The molar mass of K is O2. Since the stoichiometry has a ration of 4 mol K to mol O2 , K will be the limiting reactant when mass of K is much greater than than times the mass of O2 - 3 much lower than four five 1 comparable to 6 less three 2 more two six

Chemistry: An Atoms First Approach

2nd Edition

ISBN:9781305079243

Author:Steven S. Zumdahl, Susan A. Zumdahl

Publisher:Steven S. Zumdahl, Susan A. Zumdahl

Chapter5: Stoichiometry

Section: Chapter Questions

Problem 9RQ: Consider the following mixture of SO2(g) and O2(g). If SO2(g) and O2(g) react to form SO3(g), draw a...

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Question

Consider the following reaction:

4K(s)+O2(g)→2K2O(s)4K(s)+O2(g)→2K2O(s)The molar mass of KK is 39.10 gmol−1gmol−1 and that of O2O2 is 32.00 gmol−1gmol−1.

The limiting reactant is 1.5 g K, 0.38 g O2

I need help with the "explain your reasoning"

5
The molar mass of K is
O2. Since the stoichiometry has a ration of
4
mol K to
mol 02, K will be the limiting reactant when mass of K is
much greater than
than
times the mass of O2 .
much lower than
four
five
1
comparable to
6
less
three
2
more
two
six

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