Im stuck on these few questions that seems to be getting me no where :
-Determine the number of mmoles of HCl that did not react with the anatacid.
c HCl = 0.1812
c NaOH = 0.1511
volume of HCl added : 75
volume of NaOH added: 29.23,19.58,33.3
- mmoles of HCl neutralized by the tablet?
- mmoles of H+ neutralized per gram of antacid(mmole/g)
- Average #mmoles of H+ neutralized per gram of anatacid
To determine the number of mmol of HCl that did not react with the antacid, you'll need to find out the number of mmol of HCl that did react first.
To calculate the number of mmol of HCl that reacted with the antacid, you can use the following equation:
mmol HCl reacted = c HCl * V HCl
Where:
- c HCl is the concentration of HCl in mol/L (given as 0.1812 in your question)
- V HCl is the volume of HCl added in L (given as 75 in your question)
Plugging in the values, we get:
mmol HCl reacted = 0.1812 mol/L * 75 L
Now, you mentioned the volume of NaOH added, but we need to know the concentration of NaOH as well to determine the reaction stoichiometry. The balanced equation for the reaction between HCl and NaOH is:
HCl + NaOH → NaCl + H2O
If you can provide the concentration of NaOH, we can continue with the calculations.