I have a question regarding solution for problem 143 chapter 3 from McMurry Fay

Question

I have a question regarding solution for problem 143 chapter 3 from McMurry Fay Chemistry sixth edition text book. In step 2/4 there is a section where the solution divides (98 * .353)/80 I understand that 98 is the grams of H2SO4 per mole and .353 is the grams of NaOH used to neautrilize H2SO4, but I dont understand where the 80 comes from. Can you please exaplain where they 80 comes from. The question in case you cant access the specific sixth edition text book is:

A 1.268 g sample of a metal carbonate (MCO3) was treated with 100.00 mL of .1083 M sulfuric acid (H2SO4), yielding CO2 gas and an aqueous solution of the metal sulfate (MS)4). The solution was bioled to remove all the disolved CO2 and was then titrated with .1241 M NaOH. A 71.02 mL volume of NaOH was required to neutralize the excess H2SO4. (a) what is the identiy of the metal M? (b) how many liters of CO2 gas were produced if the density of CO2 is 1.799 g/L?

Explanation / Answer

moles H2SO4 = 0.1000 L x 0.1083 M = 0.01083
moles NaOH = 0.07102 L x 0.1241 M = 0.008814

H2SO4 + 2 NaOH = Na2SO4 + 2 H2O
moles H2SO4 = 0.008814 /2=0.004407

moles H2SO4 used to react with MCO3 = 0.01083 - 0.004407 = 0.006423

MCO3 + H2SO4 = H2O + CO2 + MSO4
moles MCO3 = 0.006423

molar mass MCO3 = 1.268 g/ 0.006423 mol=197.4 g/mol

197.4 - ( 12.011 + 3 x 15.999)=137.4 g/mol = molar mass M

M is barium

moles CO2 = 0..006423
mass CO2 = 0.006423 x 44.009 g/mol=0.2827 g
V =0.2827 / 1.799 = 0.157 L

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