Question

Ferrous sulphates are salts with the formula - $FeS{O}_4.x{H}_{2}O$. Most common of these compounds is the heptahydrate - $FeS{O}_4.7H_{2}O$. When we heat this compound to ${300}^{\circ }C$, we directly get the anhydrous Iron (II) Sulphate - True or False?

• A. True
• B. Flase

Answer 1

The correct option is A True

$FeS{O}_4.7H_{2}O\left(s\right)\stackrel{{300}^{\circ }C}{\to }FeS{O}_4+7H_{2}O\left(g\right)$

Do note that all the molecules of water need not become volatile in a single step. For instance,
$CuS{O}_4.5H_{2}O\left(s\right)\stackrel{Exposuretosunlight;efflorescence}{\to }CuS{O}_4.3H_{2}O+2H_{2}O\left(g\right)$

$CuS{O}_4.3H_{2}O\left(s\right)\stackrel{{100}^{\circ }C}{\to }CuS{O}_4.H_{2}O\left(s\right)+2H_{2}O\left(g\right)$

$CuS{O}_4.H_{2}O\left(s\right)\stackrel{{230}^{\circ }C}{\to }CuS{O}_4\left(s\right)+H_{2}O\left(g\right)$

Coming back to the original question, if we heat $FeS{O}_4.7H_{2}O$ (s) at very high temperatures,
$2FeS{O}_4\left(s\right)\stackrel{\Delta }{\to }F{e}_2{O}_3\left(s\right)+S{O}_2\left(g\right)+S{O}_3\left(g\right)$