Question

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

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

If 63 g of ammonium dichromate decomposes, calculate the mass of chromium [III] oxide formed at the same time.

Answer 1

Step 1: Write the chemical equation

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

Ammonium dichromate Nitrogen Chromium(III) oxide water

Given:

The relative molecular mass of solid ammonium dichromate = 252

Step 2: Find the mass

Molecular mass of ${\left({\mathrm{NH}}_4\right)}_2{\mathrm{Cr}}_2{\mathrm{O}}_7$

N=14 H=1 Cr=51 O=16

= 2 atoms of N + 8 atoms of H+ 1 atom of Cr + 7 atoms of O

$=2x14+8x1+52x2+7x16$

$=28+8+104+112$

$=252g/mol$

The molecular mass of ${\mathrm{Cr}}_2{\mathrm{O}}_3$

=2 atoms of Cr + 3 atoms of O

$=2x52+3x16$

$=104+48=152g/mol$

So, According to the equation,

252 g of ${\left({\mathrm{NH}}_4\right)}_2{\mathrm{CrO}}_7$ gives 152 g of ${\mathrm{Cr}}_2{\mathrm{O}}_3$

63 g of ${\left({\mathrm{NH}}_4\right)}_2{\mathrm{CrO}}_7$ will give = $152x\frac{63}{252}$ g of ${\mathrm{Cr}}_2{\mathrm{O}}_3$

= 38 g of ${\mathrm{Cr}}_2{\mathrm{O}}_3$

Step 3: Result

Hence, the mass of ${\mathrm{Cr}}_2{\mathrm{O}}_3$ formed is 38 g.