1.3188g of antacid is weighed and mixed with 75.00 mL of excess 0.1746 M HCl. The excess wcid required 27.20 mL of 0.09767 M NaOH for back titration. Calculate the amount of CaCO3 in the tablet.

The amount of CaCO₃ in the tablet : 0.5219 g

Reaction

1. CaCO₃(in antacid) + 2HCl⇒ CaCl₂+H₂CO₃

2. Titration ⇒ HCl+NaOH⇒NaCl+H₂O

For titration :

M₁V₁.n₁=M₂V₂n₂(n=acid/base valence⇒NaOH/HCl=1)

mol HCl=mol NaOH = excess HCl

mol HCl

Moles of HCl used (reacted with CaCO₃) :

initial HCl - excess HCl =

From reaction 1 :

mol HCl : mol CaCO₃ = 2 : 1

mass of CaCO₃ :

answer: the value of enthalpy change will be -152.80 kj.

solution:

pressure = 1atm , temperature = = 298 k

given that pressure and temperature are same in both gases , the number of moles of both gases can be determined by the ideal gas equation.

according to reaction 3 moles no are reacting with 1 mol of ,then 0.49 mol of no will react with moles of ozone that is 0.16 moles of ozone. since, no is limiting reagent and is an excessive reagent.

enthalpy change for the reaction between the given volumes of the gases is :

it’s previous models   because when i picked that son of a gun’s answer it was wrong look for previous models because they might make it a different letter