Question

Solid ammonium dichromate (relative molecular mass 252) decomposes as under:

${\left({\mathrm{NH}}_4\right)}_2{\mathrm{Cr}}_2{\mathrm{O}}_7\to {\mathrm{N}}_2+{\mathrm{Cr}}_2{\mathrm{O}}_3+4{\mathrm{H}}_2\mathrm{O}$

(I) If 63 g of ammonium dichromate decomposes, what will be the loss of the mass?

Answer 1

Step 1: Write the chemical equation

$$\begin{gathered} {\left({\mathrm{NH}}_4\right)}_2{\mathrm{Cr}}_2{\mathrm{O}}_7\left(\mathrm{s}\right)\to {\mathrm{N}}_2\left(\mathrm{g}\right)+{\mathrm{Cr}}_2{\mathrm{O}}_3\left(\mathrm{s}\right)+4{\mathrm{H}}_2\mathrm{O}\left(\mathrm{l}\right) \\ \mathrm{Ammonium}\mathrm{dichromate}\mathrm{Nitrogen}\mathrm{Chromium}\mathrm{oxide}\mathrm{Water} \end{gathered}$$

According to the above equation,

1 mol of ammonium dichromate would give 4 mols of H₂O.

Step 2: Find the molar mass

The molar mass of ammonium dichromate

$=2x(14+4)+2x52+7x16$

$=252\mathrm{g}/\mathrm{mol}$

The number of moles of ammonium dichromate heated = $\frac{63}{252}$$=0.25\mathrm{mol}$

Therefore, a number of moles of water evaporated =$4\mathrm{x}0.25\mathrm{mol}=1\mathrm{mol}$.

The mass of water evaporated $=1\mathrm{mol}\mathrm{x}18\mathrm{g}\mathrm{per}\mathrm{mol}=18\mathrm{g}.$

Therefore, 18 grams of water is evaporated.

Step 3: Find the loss of mass

Also, produced ${\mathrm{N}}_2$ also is released into the atmosphere.

Number of ${\mathrm{N}}_2$ moles produced $1\mathrm{x}0.25\mathrm{mol}=0.25\mathrm{mol}$

Mass of ${\mathrm{N}}_2$released $(14+14)\mathrm{x}0.25\mathrm{g}=7\mathrm{g}$

Therefore, total mass loss $=18+7=25\mathrm{g}.$

Hence the total mass loss = $25\mathrm{g}$