Question

From the equation:

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

Calculate the volume of nitrogen at STP evolved when $63\mathrm{g}$ of ammonium dichromate is heated.

Answer 1

Step 1: Given information

• The given reaction is as follows:

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

• Mass of ammonium dichromate $\left({\left({\mathrm{NH}}_4\right)}_2{\mathrm{Cr}}_2{\mathrm{O}}_7\right)=63\mathrm{g}$

Step 2: Balanced chemical reaction

• The balanced chemical reaction is as follows:

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

$\begin{array}{rcl}\mathrm{Molar}\mathrm{mass}\mathrm{of}{\left({\mathrm{NH}}_4\right)}_2{\mathrm{Cr}}_2{\mathrm{O}}_7& =& 2\times \left[14\mathrm{g}+\left(1\mathrm{g}\times 4\right)\right]+\left(52\mathrm{g}\times 2\right)+\left(16\mathrm{g}\times 7\right)\\ & =& 252\mathrm{g}\end{array}$

Volume of 1 mole of gas at STP is $22.4{\mathrm{dm}}^3$.

Step 3: Calculate the volume of nitrogen

• $252\mathrm{g}$ of ammonium dichromate evolved 1 mole or $22.4{\mathrm{dm}}^3$ of nitrogen at STP (Standard temperature and pressure).
• Thus, $63\mathrm{g}$ of ammonium dichromate will evolve $\frac{22.4{\mathrm{dm}}^3}{252\mathrm{g}}\times 63\mathrm{g}=5.6{\mathrm{dm}}^3$ of nitrogen.

Therefore, $5.6{\mathrm{dm}}^3$ of nitrogen is evolved at STP.