The balanced equation for the reaction between NaOH and HCl is as follows
NaOH + HCl ---> NaCl + H₂O
stoichiometry of NaOH to HCl is 1:1
number of NaOH moles reacted - 0.1 mol/L x 0.054 L = 0.0054 mol
number of HCl moles reacted = number of NaOH moles reacted
since the molar ratio of acid to base is 1:1
therefore number of HCl moles reacted - 0.0054 mol
the number of moles of HCl in 125 mL - 0.0054 mol
therefore number of HCl moles in 1000 mL - 0.0054 mol / 125 mL x 1000 mL
number of moles of solute in 1000 mL is known as the molarity concentration
concentration of HCl is 0.0432 M
The question is incomplete, here is the complete question:
What volume (mL) of the partially neutralized stomach acid having concentration 2M was neutralized by 0.01 M NaOH during the titration? (portion of 25.00 mL NaOH sample was used; this was the HCl remaining after the antacid tablet did it's job)
<u>Answer:</u> The volume of HCl neutralized is 0.125 mL
<u>Explanation:</u>
To calculate the concentration of acid, we use the equation given by neutralization reaction:
where,
are the n-factor, molarity and volume of acid which is HCl (Stomach acid)
are the n-factor, molarity and volume of base which is NaOH.
We are given:
Putting values in above equation, we get:
Hence, the volume of HCl neutralized is 0.125 mL
The molar mass of CO2 is 44 grams per mole.
165 grams / 44 grams per mole of CO2 = 3.75 moles CO2
Using Avogadro’s law where 1 mole of substance equals
6.023 x 10^23 molecules
3.75 moles CO2 (6.023 x 10^23 molecules /mole) = 2.26 x 10^24 molecules CO2
Answer: Poing Reyes
Explanation: the other options are incorrect as they have been used as astronomical observatory
