50 gm of 109% oleum is mixed with 50 gm of another 118% oleum. Calculate the maximum weight of H2SO 4${\mathrm{H}}_2{\mathrm{SO}}_4$ which can be obtained from the resulting mixture. (Fill the OMR after multiplying with 10)

Oleum = ${\mathrm{SO}}_3+{\mathrm{H}}_2\mathrm{O}\to {\mathrm{H}}_2{\mathrm{SO}}_4$

100g oleum + 9g ${\mathrm{H}}_2\mathrm{O}$

100g oleum $S{O}_3\to {40}_y$

${\mathrm{H}}_2{\mathrm{SO}}_4\to 100-40=60g$

moles of  ${\mathrm{SO}}_3$= moles of ${\mathrm{H}}_2\mathrm{O}$

$\frac{x}{80}=\frac{9}{18}$

$\mathrm{x}=\frac{9}{18}\times 80=40\mathrm{y}$

50 oleum ${\mathrm{SO}}_3\to 4{\mathrm{Og}}_{ }$

$H_2{\mathrm{SO}}_4\to 10y$

Total ${\mathrm{SO}}_3$ = $40+20=60\mathrm{g}$

Total ${\mathrm{H}}_2{\mathrm{SO}}_4$ = $30+10=40y$

${\mathrm{SO}}_3+{\mathrm{H}}_2\mathrm{O}\to {\mathrm{H}}_2{\mathrm{SO}}_4$

moles of ${\mathrm{SO}}_3$ = moles of ${\mathrm{H}}_2{\mathrm{SO}}_4$

Wt. of ${\mathrm{H}}_2{\mathrm{SO}}_4$ = $\frac{50\times 109+50\times 118}{100}=113.5\mathrm{\%}$