Question

Suppose you start with 1.80 g of mgco3·5h2o, after complete dehydration how many grams of the sample do you have remaining?

Answer 1

The  number of  grams of the sample that remained  after  hydration is 0.87  grams


                calculation
calculate the molar  mass of MgCO3. 5 H2O  which is

24 + 12+ (3 x16) + ( 5  x18)  =  174 g/mol

we  well  know  after  hydration  water evaporate leaving behind   MgCO3
therefore the  mass  remained  is  for  MgCO3

find  the   molar  mass of  MgCO3 =  24 +12 +( 3 x16)  = 84  g/mol

the  mass  remained is therefore

 =  molar  mass of MgCO3 /  molar mass of MgCO3.5H2O  x 1.80 g

=  84g/mol / 174 g/mol  x1.8g = 0.87  grams