Consider the following reaction: 4K(s)+02(g) →2K2O(s)4K(s)+02(g) →2K20(s) The molar mass of KK is 39.10 gmol-1gmol-1 and that of 0202 is 32.00 gmol-1gmol-1. The limiting reactant is 1.5 g K, 0.38 g 02 I need help with the "explain your reasoning" 5 4 much greater than 3 much lower than four five 1 comparable to 6 less three 2 more two six The molar mass of K is mol K to than O2. Since the stoichiometry has a ration of mol O₂, K will be the limiting reactant when mass of K is times the mass of O₂.

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Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
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Chapter1: Chemical Foundations
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Consider the following reaction:
4K(s)+02(g) →2K2O(s)4K(s)+O2(g)→2K2O(s)The
molar mass of KK is 39.10 gmol-1gmol-1 and that of
0202 is 32.00 gmol-1gmol-1.
The limiting reactant is 1.5 g K, 0.38 g 02
I need help with the "explain your reasoning"
K
5
4
much greater than
3
much lower than
four
five
1
comparable to
6
less
three
2
more
two
six
The molar mass of K is
mol K to
than
O2. Since the stoichiometry has a ration of
mol O₂, K will be the limiting reactant when mass of K is
times the mass of 0₂.
38
Transcribed Image Text:Consider the following reaction: 4K(s)+02(g) →2K2O(s)4K(s)+O2(g)→2K2O(s)The molar mass of KK is 39.10 gmol-1gmol-1 and that of 0202 is 32.00 gmol-1gmol-1. The limiting reactant is 1.5 g K, 0.38 g 02 I need help with the "explain your reasoning" K 5 4 much greater than 3 much lower than four five 1 comparable to 6 less three 2 more two six The molar mass of K is mol K to than O2. Since the stoichiometry has a ration of mol O₂, K will be the limiting reactant when mass of K is times the mass of 0₂. 38
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