For every 1 mole of C6H12O6, you need 6 moles of water. Multiply the 2.5 moles you are trying to make by the 6 of water you need, and 4) 15 is your answer.
The limiting reagent when 5 g of NaOH and 4.4 g CO₂ allowed to react will be NaOH
<h3>What is Limiting reagent ?</h3>
The limiting reactant (or limiting reagent) is the reactant that gets consumed first in a chemical reaction and therefore limits how much product can be formed.
Given chemical equation in balanced form ;
2NaOH(s) + CO₂(g) → Na₂CO₃(s) + H₂O(l).
According to the Chemical equation ;
- The limiting reagent when 5 g of NaOH and 4.4 g CO₂ allowed to react will be NaOH
If 44 g CO₂ requires 80 g of NaOH, therefore, 4.4 g CO₂ will require atleast 8 g of NaOH.
But the available quantity is 5 g NaOH. thus, NaOH is the Limiting reagent.
- 6.625 g of Na₂CO₃ are expected to be produced 5.0 g of NaOH and 4.4 g of CO₂ are allowed to react
As 80 g NaOH produces 106 g of Na₂CO₃.
Therefore 5 g NaoH will produce ;
106 / 80 x 5 = 6.625 g
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Option “A” is the crest of the wave because it’s the maximum value of upward displacement within a cycle.
Explanation:
BIOMEDICAL IMPORTANCE
In addition to serving as precursors of nucleic acids, purine and pyrimidine nucleotides participate in metabolic functions as diverse as energy metabolism, protein synthesis, regulation of enzyme activity, and signal transduction.
B and C are in excess so amount of E will be determined by A.
Amount of product is determined by limiting reagents - Always.
Hence 6 moles of E will be formed.
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