Hi,
14H
15B
16J
17A
I hope this helps. If you’d like further explanation please let me know.
Answer:
KO is the limiting reactant.
0.11 mol O₂ will be produced.
Explanation:
4 KO₂ + 2 H₂O ⇒ 4 KOH + 3 O₂
Find the limiting reagent by dividing the moles of the reactant by the coefficient in the equation.
(0.15 mol KO₂)/4 = 0.0375
(0.10 mol H₂O)/2 = 0.05
KO₂ is the limiting reagent.
The amount of product produced depends on the limiting reagent. To find how much is produced, take moles of limiting reagent and multiply it by the ratio of reagent to product. You can find the ratio by looking at the equation. For every 4 moles of KO₂, 3 moles of O₂ are produced.
0.15 mol KO₂ (3 mol O₂)/(4 mol KO₂) = 0.1125 mol O₂
0.11 mol O are produced.
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Did you get the answer bc it would rly help me out thanks