Calculate the number of moles of HCl that reacted. Molarity of HCI = 0.10 mol dm3. Now, 1 dm³ is equal to 1 L, so, molarity of HCI is 0.10 mol/L Volume of HCI taken = 25.00 cm³ = 25.00 x 103 L [1 cm³ = 10° L] Moles of HCI reacted = molarity * volume in L = 0.10 mol/L * (25.00 x 10° L) = 2.500 x 103 mol The moles of HCI that reacted is 2.5 x 103 mol. By reference to the reacting ratio of HCl and NaOH, deduce the number of moles of NaOH required to reach equivalence.
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By reference to the reacting ratio of HCL and NaOH, deduce the number of moles of NaOH required to reach Equivalence.
Btw I did the first part of the question so this is a follow up.
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Analysis
Calculate the number of moles of HCl that reacted.
Molarity of HCI = 0.10 mol dm3.
Now, 1 dm³ is equal to 1 L, so, molarity of HCI is 0.10 mol/L
Volume of HCI taken = 25.00 cm3
= 25.00 x 103L
[1 cm³ = 10° L]
%3D
%3D
Moles of HCI reacted =
molarity * volume in L
%3D
= 0.10 mol/L * (25.00 x 10° L)
= 2.500 x 103 mol
The moles of HCI that reacted is 2.5 x 103 mol.
By reference to the reacting ratio of HCl and NaOH, deduce the number of moles of NaOH required to reach
equivalence.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F2433ae66-1fb6-435c-a03a-a02f5b5a579e%2Fe93c3bf6-721b-4f3d-894b-3e55ed070e40%2Ft5mvrp8_processed.jpeg&w=3840&q=75)
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