Question

How many moles of caco3 are there in an antacid tablet containing 0.515 g caco3?

Answer 1

The  moles  of  CaCO3  which  are  there  in  antacid  tablet  that  contain  0.515g  CaCO3  is  calculated  as  follows

moles  =mass/molar  mass

the  molar  mass  of  CaCO3 =  ( 40  x1)+  (12  x  1)  + (16 x  3)=  100  g/mol

moles  is  therefore=  0.515g/100 g/mol=  5.15  x10^-3  moles

Answer 2

Answer:

$0.00515molCaCO3$

Explanation:

Hello,

By following the down below simple mass-mole relationship, the requested moles are computed, considering that the calcium carbonate has a molecular mass of 100g/mol:

$M_{CaCO3}=40+12+16*3=100g/mol\\\\n_{CaCO3}=0.515gCaCO3*\frac{1molCaCO3}{100gCaCO3}=0.00515molCaCO3$

Best regards.