Question

Complete and balance the equations for the following reactions.

(a) $\mathit{L}\mathit{i}\mathbf{\left(}\mathbf{s}\mathbf{\right)}\mathbf{+}{\mathit{N}}_{\mathbf{2}}\mathbf{\left(}\mathbf{g}\mathbf{\right)}\mathbf{\to }$

(b)$\mathit{R}\mathit{b}\left(s\right)\mathbf{+}\mathit{S}\left(s\right)\mathbf{\to }$

(c)$\mathit{C}\mathit{s}\left(s\right)\mathbf{+}{\mathit{H}}_{\mathbf{2}}\mathit{O}\left(l\right)\mathbf{\to }$

(d) $\mathit{N}\mathit{a}\left(s\right)\mathbf{+}\mathit{C}{\mathit{l}}_{\mathbf{2}}\left(g\right)\mathbf{\to }$

Short answer

Answer:

(a) $6Li\left(s\right)+3N_2\left(g\right)\to 2L{i}_{3}N\left(s\right)$

(b) $\mathit{2}Rb\mathit{\left(}\mathit{s}\mathit{\right)}\mathit{+}S\mathit{\left(}\mathit{s}\mathit{\right)}\mathit{\to }R{b}_{\mathit{2}}S\left(s\right)$

(c) $\mathit{2}Cs\mathit{\left(}\mathit{s}\mathit{\right)}\mathit{+}\mathit{2}{H}_{\mathit{2}}O\mathit{\left(}\mathit{l}\mathit{\right)}\mathit{\to }{H}_{\mathit{2}}\left(g\right)\mathit{+}\mathit{2}C{s}^{\mathit{+}}\left(aq\right)\mathit{+}\mathit{2}O{H}^{\mathit{-}}\left(aq\right)\mathit{/}\mathit{2}CsOH\left(aq\right)\mathit{+}{H}_{\mathit{2}}\left(g\right)$

(d) $\mathit{2}Na\mathit{\left(}\mathit{s}\mathit{\right)}\mathit{+}C{l}_{\mathit{2}}\mathit{\left(}\mathit{g}\mathit{\right)}\mathit{\to }\mathit{2}NaCl$

Step-by-step solution

Determining the products formed on reaction of Li(s) and N2(g) and balancing it.

6 moles of lithium react with 3 moles of nitrogen to produce 2 moles of Lithium nitride. Here we start by balancing the nitrogen atoms first since the Li atom is a single element.

Put a “2” as the coefficient on the $L{i}_{\mathit{3}}N$. This will balance the N-atoms.

Change the coefficient in front of the Li to finish balancing the equation.

Determining the products formed on reaction of Rb(s) and S(s) and balancing it.

2 moles of rubidium react with 1 mole of sulphur to produce 1 mole of Rubidium sulfide. Here we start by balancing the sulphur atom first. Since, the number of sulphur atoms are same on the LHS and RHS so the coefficient is 1.

Put a “2” as the coefficient on rubidium on the reactant’s side to balance out the number of rubidium atoms.

Determining the products formed on reaction of  Cs(s)+H2O(l) and balancing it.

2 moles of caesium react with 2 moles of water to produce 2 moles of Caesium hydroxide along with hydrogen gas.

To balance the number of hydrogens on the reactant side, 2 is multiplied to $CsOH$as the coefficient.

To balance the number of oxygens on the product side, 2 is multiplied to$H_{\mathit{2}}O$as the coefficient.

To balance the number of caesium atoms on the product side, 2 is multiplied by caesium as the coefficient.

Determining the products formed on reaction of Na and Cl2 and balancing it.

2 moles of sodium atom react with 2 moles of $C{l}_{\mathit{2}}$to produce 2 moles of $NaCl$.

Since there are 2 chlorine atoms on the left side, there need to be two chlorine atoms on the right side. So put a coefficient of two in front of the formula $NaCl$.

Now there are two sodium atoms and two chlorine atoms on the right side. However, the left side has only one sodium atom. So put a coefficient of 2 in front of the $Na$atom.

Now there are two sodium and two chlorine atoms on both sides of the equation and hence it is balanced.