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

Answer 2

Dehydration is the process of loss of water from the sample. The mass of sample remains after dehydration is 0.87 grams.

What is the mass of the sample after dehydration?

The magnesium carbonate pentahydrate is a 5 water molecules containing compound.

The molar mass of magnesium carbonate pentahydrate is 174 g/mol. The mass of water in the compound is 90 grams.

After the loss of water from 1 mole that is 174 grams magnesium carbonate pentahydrate, the mass remains is:

$\rm Magnesium \;carbonate\;mass=Magensium\;carbonate\;pentahydrate\;mass-Water\;mass\\Magnesium \;carbonate\;mass=174-90\;g\\Magnesium \;carbonate\;mass=84\;g$

The mass of sample remained in the dehydration of a mole of sample is 84 grams.

The mass of sample remains with the dehydration of 1.8 grams of sample is given as:

$\rm 174\;g\;MgCO_.5H_2O=84\;g\;MgCO_3\\\\1.80\;g\;MgCO_3.5H_2O=\dfrac{84}{174}\;\times\;1.80\;g\;MgCO_3\\\\ 1.80\;g\;MgCO_3.5H_2O=0.87\;g\;MgCO_3$

The mass of sample remained after dehydration of 1.80 g, Magnesium carbonate pentahydrate is 0.87 grams.

Learn more about dehydration, here;

https://brainly.com/question/16394592