Calcium carbonate reacts with hydrobromic acid to form calcium bromide, carbon dioxide, and water according to the reaction below. what is the molarity of the hydrobromic acid if 250.0 ml of it reacts with 5.64 grams of calcium carbonate?

The balanced equation for the above reaction is as follows
CaCO₃ + 2HBr ---> CaBr₂ + CO₂ + H₂O
stoichiometry of CaCO₃ to HBr is 1:2
number of moles of CaCO₃ reacted - 5.64 g / 100 g/mol = 0.0564 mol 
according to molar ratio
number of HBr moles - 0.0564 mol x 2 = 0.1128 mol 
number of HBr moles in 250.0 mL - 0.1128 mol
therefore number of HBr moles in 1000 mL - 0.1128 mol / 250.0 mL x 1000 mL = 0.4512 mol 
molarity of HBr - 0.4512 M

Answer;

= 3.064 g/L

Explanation;

Using the equation;

PV = nRT , where P is the pressure,. V is the volume, n is the number of moles and T is the temperature and R is the gas constant, 0.08206 L. atm. mol−1.

Number of moles is 1 since one mole has a mass equivalent to the molar mass.

Therefore; We can find the volume and thus get the density.

V = nRT/P

   = (1 × 0.08206 × 498 )/ 0.939

   = 43.52 L

But; density = mass/volume

Therefore;

Density = 133.34 / 43.52

             = 3.064 g/L