Question

Igniting $Mn{O}_2$ in air converts it quantitatively to ${Mn}_3{O}_4$. A sample of pyrolusite has $80\%$ $Mn{O}_2$, $15\%$ $Si{O}_2$ and the rest is water. The sample is heated in air to a constant mass. The $\%$ of $Mn$ in the ignited sample is (as the nearest integer)

Answer 1

On ignition, the reactions that take place are
$3Mn{O}_2\stackrel{\Delta }{\to }{Mn}_3{O}_4+O_2$
$Si{O}_2\stackrel{\Delta }{\to }Nochange$
$H_2{O}_l\stackrel{\Delta }{\to }{H}_2{O}_{\left(v\right)}\uparrow$
The sample of pyrolusite on heating left a residue of ${Mn}_3{O}_4+Si{O}_2$
$\because 3\times 87$ g $Mn{O}_2$ gives $229$ g ${Mn}_3{O}_4$
$80$ g $Mn{O}_2$ gives $229\times \frac{80}{(3\times 87)}=70.19$ g ${Mn}_3{O}_4$
Mass of residue $=$ $70.19$ g ${Mn}_3{O}_4$ + $15$ g $Si{O}_2$ $=$ $85.19$ g
$229$ g ${Mn}_3{O}_4$ has $Mn$ $=55\times 3$ g
$70.19$ g ${Mn}_3{O}_4$ has $Mn$ $=\frac{55\times 3\times 70.19}{229}$ g $=$ 50.58 g $Mn$
The residue $85.19$ g has $50.58$ g of $Mn$.
$100$ g has $Mn$ $=\frac{50.58\times 100}{85.19}=$ $59.37\%$
$\%$ $Mn$ in residue (ignited sample) $=59.37$
So, nearest integer is $59$.