Copper(ll) sulphate pentahydrate, CuSO4.5H2O is a bright blue crystal. upon heating, it decomposes to white powder copper(ll) sulphate, CuSO4 and water, H2O. given a sample of 10.0g CuSO4. 5H2O, calculate i. the amount of copper(ll) sulphate pentahydrate in mole
ii. the amount of water molecules liberated upon heating in mole
iii. the mass of anhydrous copper(ll) sulphate in gram

the correct answer is option d

explanation:

i was wrong its d because they dont use cm when calculating square roots

"you start with 15.67 g mgco3-xh2o, when heated and all the water is removed, you are left with 7.58 g mgco3 only.

therefore, you had 15.67 - 7.58 = 8.09g h2o  

moles of each :

7.58g mgco3 = ? moles = 7.58g / 83.4g/mole = 0.09moles  

8 09g h2o = ? moles = 8.09g / 18g/mole = 0.44moles  

molar ratio between the two by dividing both numbers by the smallest number :

0.09 / 0.09 = 1  

0.44 / 0.09 = 4.8  

mgco3-5h2o"

it is stable as it is in group 0 which have full outer shells of electrons