Answer:
Explanation:
mole of HCl remaining after reaction with CaCO₃
= .3 M of NaOH of 32.47 mL
= .3 x .03247 moles
= .009741 moles
Initial HCl taken = .3 x .005 moles = .0015 moles
Moles of HCl reacted with CaCO₃
= .009741 - .0015 = .008241 moles
CaCO₃   +   2HCl  =  CaCl₂  +  CO₂  +  H₂O .
1 mole     2 moles
2 moles of HCl reacts with 1 mole of  CaCO₃
.008241 moles of HCl reacts with .5 x .008241 moles of  CaCO₃
CaCO₃ reacted with HCl =  .5 x .008241 = .00412 moles
the mass (in grams) of calcium carbonate in the tablet
= .00412 x 100 = .412 grams . ( molar mass of calcium carbonate = 100 )
We have that  the mass (in grams) of calcium carbonate in the tablet  is mathematically given as
mass of  CaCO₃= 0.412 grams
Question Parameters:
Generally the equation for the mole of HCl  is mathematically given as
mole of HCl = 0.3 M of NaOH of 32.47 mL
mole of HCl= 0.3 x 0.03247 moles
mole of HCl= 0.009741 moles
Therefore
Initial HCl taken = 0.3 x 0.005 moles
Initial HCl taken= 0.0015 moles
Moles of HCl reacted with CaCO₃
M_{hcl}= 0.009741 - .0015
M_{hcl}= .008241 moles
The Chemical Equation
CaCO₃   +   2HCl  =  CaCl₂  +  CO₂  +  H₂O .
Therefore, CaCO₃ reacted with HCl is
CaCO₃ reacted with HCl=  0.5 x 0.008241
 CaCO₃ reacted with HCl= .00412 moles
Hence,the mass of  CaCO₃ is
mass of  CaCO₃ = 0.00412 x 100
mass of  CaCO₃= 0.412 grams
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