Chemistry - Stoichiometry Question with Solution | TestHub

No. of moles of ${\mathrm{O}}_2=\frac{56}{22.4}=2.5$

No. of moles of $\mathrm{C}=\frac{84}{12}=7$

No. of moles of ${\mathrm{H}}_2=\frac{12}{2}=6$

$12\mathrm{C}+11{\mathrm{H}}_2+\frac{11}{2}{\mathrm{O}}_2\to {\mathrm{C}}_{12}{\mathrm{H}}_{22}{\mathrm{O}}_{11}$

O₂ is limiting reagent

$\frac{11}{2}$ moles ${\mathrm{O}}_2$ gives $=342\mathrm{g}$ sucrose (i.e. 1 mole)

2.5 mole O₂ gives =

$\frac{342}{\frac{11}{2}}\times 2.5=155.5gm$