Question

a) Complete and balance the following reactions: [5]

(i) Zn + NaOH ------------->

(ii) Al + H2SO4 ------------->

(iii) Fe + HCl -------------->

(iv) Mg + H2O (steam) ------------>

(v) Cu + dil. HCl --------------->​

Answer 1

$\text{(i) Zn + NaOH} \longrightarrow \textsf{No reaction}$

• Reason : In Reactivity series of metals, Zinc(Zn) lies below Sodium(Na), so it can't replace sodium from its hydroxide. Thus, no reaction occur when Zinc is added to Sodium Hydroxide.

$\rm{(ii) \: \: Al + H_2SO_4} \longrightarrow \: \underline{Al_2 (SO_4)_3 + H_2\uparrow} \\ \\ \small \rm{Balanced \: \: {Eq}^{n} } \rightarrow \: { \green{{2Al + 3H_2SO_4} \longrightarrow \: Al_2 (SO_4)_3 + 3H_2\uparrow }}$

${\sf{(iii) \: \: Fe + HCl \longrightarrow \underline{FeCl_2 + H_2\uparrow} }}$

${\sf{ Balanced \: \: Eq^{n} \rightarrow \green{ Fe + 2HCl \longrightarrow FeCl_2 + H_2\uparrow}}}$

${\sf{(iv) \: \: Mg + H_2O_{(steam)} \longrightarrow \underline{MgO + H_2 \uparrow}}}$

$\: {\bf{ \bull\: \: It \: \:is \: \:already\: \: balanced.}}$

$\sf{(v) \: \: Cu + {\tiny{(dil.)}} HCl \longrightarrow \underline{CuCl_2 + H_2}} \\ \\ \sf{ Balanced \: \: Eq^{n} \rightarrow \green{Cu + 2HCl \longrightarrow CuCl_2 + H_2 \uparrow}}$