Question

From the equation,$\mathrm{C}+2{\mathrm{H}}_2{\mathrm{SO}}_4\to {\mathrm{CO}}_2+2{\mathrm{H}}_2\mathrm{O}+2{\mathrm{SO}}_2$

Calculate the mass of carbon oxidised by 49g of Sulphuric acid (C-12 relative molecular mass of Sulphuric acid-98)

The volume of sulphur dioxide measured at STP liberated at the same time (volume occupied by 1 mole of a gas at STP is $22.4{\mathrm{dm}}^3$)

Answer 1

Step-1 Calculate the mass of oxidized Carbon(C)

The given equation, $C\left(g\right)+2H_{2}S{O}_4\left(aq\right)\to C{O}_{2_{}}\left(g\right)+2H_{2}O\left(l\right)+2S{O}_2\left(g\right)$

Molar mass of $H_{2}S{O}_4=196g$

$196g$ of $H_{2}S{O}_4$ will oxidized = $12g$ of Carbon

$1g$ of $H_{2}S{O}_4$ will oxidized = $\frac{12}{196}$ of Carbon

$49g$ of $H_{2}S{O}_4$ will oxidized = $\frac{12}{196}\times 49g=3g$ of Carbon

Step-2 Calculate the volume of Sulphur dioxide $S{O}_2\left(g\right)$

From the equation,$2$ moles of Sulphuric acid $\left(H_{2}S{O}_4\right)$will release $2$ moles of Sulphur dioxide $\left(S{O}_2\right)$

$196g$ of $H_{2}S{O}_4$ liberate = $=2\times 22.4LofS{O}_2$

$49g$ of $H_{2}S{O}_4$ liberate = $\frac{\left(2\times 22.4\times 49\right)}{196}=11.2\mathrm{L}ofS{O}_2$

Hence, the mass of Carbon dioxide = $3g$

the volume of Sulphur dioxide at STP= $11.2L$