Answer
$CuSO_{4}.5H_{2}O$ When Heated to $110 ^{\circ}C$ it gives $CuSO_{4}.H_{2}O$. When it is heated above $150 ^{\circ}C$ it loses all water molecule and form $CuSO_{4}$.
Molar mass of $CuSO_{4}.5H_{2}O$ = 249.7 g/mol
Molar mass of $CuSO_{4}.H_{2}O$ = 177.6 g/mol
Molar mass of $CuSO_{4}$ = 159.6 g
a) How many grams of $CuSO_{4} .H_{2}O$ could be obtained by heating 495 g of $CuSO_{4}.5H_{2}O$ to $110 ^{\circ}C$
One mole of $CuSO_{4} .5H_{2}O$ gives 1 mole of $CuSO_{4} .H_{2}O$ on heating at $110 ^{\circ}C$.
No of moles of $CuSO_{4} .5H_{2}O$ = 495 g / 249.7 g = 1.982 mole
So. 1.982 moles of $CuSO_{4} .5H_{2}O$ gives 1.982 mole of $CuSO_{4} .H_{2}O$ on heating at $110 ^{\circ}C$.
Mass of $CuSO_{4} .H_{2}O$ = ( Molar mass of $CuSO_{4} .H_{2}O$ $ \times$ Moles of $CuSO_{4} .H_{2}O$ )
Mass of $CuSO_{4} .H_{2}O$ = (177.6 g/mol $\times$ 1.982 mol)
= 352 g
b) How many grams of anhydrous $CuSO_{4}$ could be obtained by heating 463 g of $CuSO_{4}.5H_{2}O$ to $180 ^{\circ}C$.
One mole of $CuSO_{4} .5H_{2}O$ gives 1 mole of $CuSO_{4} $ on heating at $180 ^{\circ}C$.
No of moles of $CuSO_{4} .5H_{2}O$ = 463 g / 249.7 g = 1.854 mole
Mass of $CuSO_{4} $ = (159.6 g/mol $\times$ 1.854 mol)
=295.89 g
$\approx$ 296 g
Work Step by Step
$CuSO_{4}.5H_{2}O$ When Heated to $110 ^{\circ}C$ it gives $CuSO_{4}.H_{2}O$. When it is heated above $150 ^{\circ}C$ it loses all water molecule and form $CuSO_{4}$.
Molar mass of $CuSO_{4}.5H_{2}O$ = 249.7 g/mol
Molar mass of $CuSO_{4}.H_{2}O$ = 177.6 g/mol
Molar mass of $CuSO_{4}$ = 159.6 g
No of moles =( Given Mass / Malar Mass )
a) How many grams of $CuSO_{4} .H_{2}O$ could be obtained by heating 495 g of $CuSO_{4}.5H_{2}O$ to $110 ^{\circ}C$
One mole of $CuSO_{4} .5H_{2}O$ gives 1 mole of $CuSO_{4} .H_{2}O$ on heating at $110 ^{\circ}C$.
No of moles of $CuSO_{4} .5H_{2}O$ = 495 g / 249.7 g = 1.982 mole
So, 1.982 moles of $CuSO_{4} .5H_{2}O$ gives 1.982 mole of $CuSO_{4} .H_{2}O$ on heating at $110 ^{\circ}C$.
Mass of $CuSO_{4} .H_{2}O$ = ( Molar mass of $CuSO_{4} .H_{2}O$ $ \times$ Moles of $CuSO_{4} .H_{2}O$ )
Mass of $CuSO_{4} .H_{2}O$ = (177.6 g/mol $\times$ 1.982 mol) = 352 g
b) How many grams of anhydrous $CuSO_{4}$ could be obtained by heating 463 g of $CuSO_{4}.5H_{2}O$ to $180 ^{\circ}C$.
One mole of $CuSO_{4} .5H_{2}O$ gives 1 mole of $CuSO_{4} $ on heating at $180 ^{\circ}C$.
No of moles of $CuSO_{4} .5H_{2}O$ = 463 g / 249.7 g = 1.854 mole
Mass of $CuSO_{4} $ = (159.6 g/mol $\times$ 1.854 mol)=295.89 g $\approx$ 296 g
