Question

One of relatively few reactions that take place directly between two solids at room temperatureis

$$\begin{gathered} \mathbf{Ba}{\left(\mathbf{OH}\right)}_{\mathbf{2}}\mathbf{·}\mathbf{8}{\mathbf{H}}_{\mathbf{2}}\mathbf{O}\left(\mathbf{s}\right)\mathbf{+}{\mathbf{NH}}_{\mathbf{4}}\mathbf{SCN}\left(\mathbf{s}\right)\to \mathbf{Ba}{\left(\mathbf{SCN}\right)}_{\mathbf{2}}\left(\mathbf{s}\right)\mathbf{+}{\mathbf{H}}_{\mathbf{2}}\mathbf{O}\left(\mathbf{l}\right)\mathbf{+} \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}{\mathbf{NH}}_{\mathbf{3}}\left(\mathbf{g}\right) \end{gathered}$$

In this equation, the localid="1662172238669" $\mathbf{8}{\mathbf{H}}_{\mathbf{2}}\mathbf{O}$in$\mathbf{Ba}{\left(\mathbf{OH}\right)}_{\mathbf{2}}\mathbf{·}\mathbf{8}{\mathbf{H}}_{\mathbf{2}}\mathbf{O}\left(\mathbf{s}\right)$indicates the presence of eight water molecules. This compound is called barium hydroxide octahydrate.
a. Balance the equation.
b. What mass of ammonium thiocyanate $\mathbf{\left(}{\mathbf{\text{NH}}}_{\mathbf{4}}\mathbf{\text{SCN}}\mathbf{\right)}$must be used if it is to react completely with $\mathbf{6}\mathbf{.}\mathbf{5}\mathbf{}\mathbf{\text{g}}$barium hydroxide octahydrate?

Short answer

a) The balanced equation is

b) The mass of ${\mathbf{\text{NH}}}_{\mathbf{4}}\mathbf{\text{SCN}}$used to react completed withlocalid="1662172245221" $6.5\text{g}$barium hydroxide octahydrate is localid="1662172249074" $\mathbf{3}\mathbf{.}\mathbf{2}\mathbf{}{\mathbf{\text{gNH}}}_{\mathbf{4}}\mathbf{\text{SCN}}$

Step-by-step solution

Definition

A balanced chemical equation is one in which the number of atoms present on the reactant side of the equation equals the number of atoms present on the product side of the equation.

Balancing Oxygen atoms

a)

The given equation is as follows:

$$\begin{gathered} \mathrm{Ba}{\left(\mathrm{OH}\right)}_2·8{\mathrm{H}}_2\mathrm{O}\left(\mathrm{s}\right)+{\mathrm{NH}}_4\mathrm{SCN}\left(\mathrm{s}\right)\to \mathrm{Ba}{\left(\mathrm{SCN}\right)}_2\left(\mathrm{s}\right)+{\mathrm{H}}_{2}O\left(l\right) \\ +{\mathrm{NH}}_3\left(\mathrm{g}\right) \end{gathered}$$

On reactants, the number of $\text{O}$atoms is ten, whereas, on products, it is one. As a result, giving ${\text{H}}_2\text{O}$a coefficient of 10 will bring the $\text{O}$atoms into balance, as seen below:

$$\begin{gathered} \mathrm{Ba}{\left(\mathrm{OH}\right)}_2·8{\mathrm{H}}_2\mathrm{O}\left(\mathrm{s}\right)+{\mathrm{NH}}_4\mathrm{SCN}\left(\mathrm{s}\right)\to \mathrm{Ba}{\left(\mathrm{SCN}\right)}_2\left(\mathrm{s}\right)+10{\mathrm{H}}_{2}O\left(l\right) \\ +{\mathrm{NH}}_3\left(\mathrm{g}\right) \end{gathered}$$

Balancing SCN  group atoms

Similarly,the $\text{SCN}$group can be balanced by assigning a coefficient 2to$\left({\mathrm{NH}}_4\mathrm{SCN}\right)$, resulting in a total of 2 $\mathrm{SCN}$groups on each side of the reaction, as shown below:

$$\begin{gathered} \mathrm{Ba}{\left(\mathrm{OH}\right)}_2·8{\mathrm{H}}_2\mathrm{O}\left(\mathrm{s}\right)+2{\mathrm{NH}}_4\mathrm{SCN}\left(\mathrm{s}\right)\to \mathrm{Ba}{\left(\mathrm{SCN}\right)}_2\left(\mathrm{s}\right)+10{\mathrm{H}}_{2}O\left(l\right) \\ +{\mathrm{NH}}_3\left(\mathrm{g}\right) \end{gathered}$$

Balanced Equation

As a result, the number of$\text{N}$atoms on the reactant is reduced to two, and a coefficient of two is applied to ${\text{NH}}_3$ balancethe $\text{N}$atoms, as shown below. This also ensures that the $\text{H}$ atoms on both sides of the reaction are balanced.

Therefore the completed balanced equation is given as:

$$\begin{gathered} \mathrm{Ba}{\left(\mathrm{OH}\right)}_2·8{\mathrm{H}}_2\mathrm{O}\left(\mathrm{s}\right)+2{\mathrm{NH}}_4\mathrm{SCN}\left(\mathrm{s}\right)\to \mathrm{Ba}{\left(\mathrm{SCN}\right)}_2\left(\mathrm{s}\right)+10{\mathrm{H}}_{2}O\left(l\right) \\ +2{\mathrm{NH}}_3\left(\mathrm{g}\right) \end{gathered}$$

Determining the moles of Ba(OH)2·8H2O(s)

b)

We know that the $\mathbf{\text{molecularmassofBa}}{\left(\text{OH}\right)}_{\mathbf{2}}\mathbf{·}\mathbf{8}{\mathbf{\text{H}}}_{\mathbf{2}}\mathbf{\text{O}}\left(\text{s}\right)$

localid="1662172353766" $=316.46\raisebox{1ex}{$\text{g}$}\!\left/ \!\raisebox{-1ex}{$\text{mol}$}\right.$

Let’s calculate localid="1662172357775" $\mathbf{\text{thenumberofmolesofBa}}{\mathbf{\left(}\mathbf{\text{OH}}\mathbf{\right)}}_{\mathbf{2}}\mathbf{·}\mathbf{8}{\mathbf{\text{H}}}_{\mathbf{2}}\mathbf{\text{O}}\mathbf{\left(}\mathbf{s}\mathbf{\right)}$ in localid="1662172362474" $6.5\mathrm{g}$as follows:

localid="1662172366615" $\text{MolesofBa}{\left(\mathrm{OH}\right)}_2·8{\mathrm{H}}_2\mathrm{O}\left(\mathrm{s}\right)=\frac{\mathrm{Weight}\mathrm{of}\mathrm{Ba}{\left(\mathrm{OH}\right)}_2·8{\mathrm{H}}_2\mathrm{O}\left(\mathrm{s}\right)}{\mathrm{Molar}\mathrm{mass}\mathrm{of}\mathrm{Ba}{\left(\mathrm{OH}\right)}_2·8{\mathrm{H}}_2\mathrm{O}\left(\mathrm{s}\right)}$

localid="1662172370177" $=\frac{\text{6.6g}}{\text{315.46}{\displaystyle \raisebox{1ex}{$\text{g}$}\!\left/ \!\raisebox{-1ex}{$\text{mol}$}\right.}}$

localid="1662172374021" $=0.021\text{mol}\mathrm{Ba}{\left(\mathrm{OH}\right)}_2·8{\mathrm{H}}_2\mathrm{O}$

Determining the mass of (NH4SCN)

One mole of $\mathrm{Ba}{\left(\mathrm{OH}\right)}_2·8{\mathrm{H}}_2\mathrm{O}\left(\mathrm{s}\right)$reacts with two moles of $\left({\mathrm{NH}}_4\mathrm{SCN}\right)$, according to the balanced equation found in part (a). As a result:

$0.021\text{mol}\mathrm{Ba}{\left(\mathrm{OH}\right)}_2·8{\mathrm{H}}_2\mathrm{O}$reacts with

$0.042{\text{molNH}}_4\text{SCN}$$\left(0.021\times 2=0.042\right)$.$\left({\mathrm{NH}}_4\mathrm{SCN}\right)$has a molar mass of$76.13\raisebox{1ex}{$\text{g}$}\!\left/ \!\raisebox{-1ex}{$\text{mol}$}\right.$.

Therefore, the mass of $\left({\mathrm{NH}}_4\mathrm{SCN}\right)$needed for the reaction can be computed as follows:

${\mathbf{\text{MassofNH}}}_{\mathbf{4}}\mathbf{\text{SCN}}\mathbf{=}\mathbf{\text{numberofmoles}}\mathbf{\times }\mathbf{\text{molarmass}}$

$=0.042\times \frac{76.13g}{mol}$

$=3.20\mathrm{g}{\mathrm{NH}}_4\mathrm{SCN}$

Therefore, the mass of ${\mathbf{\text{NH}}}_{\mathbf{4}}\mathbf{\text{SCN}}$required to react completely with$6.5\mathrm{g}$barium hydroxide octahydrate would be $\mathbf{3}\mathbf{.}\mathbf{20}{\mathbf{\text{gNH}}}_{\mathbf{4}}\mathbf{\text{SCN}}$.