Question

Sulphur trioxide is prepared by the following two reactions:
$S_8\left(s\right)+8O_2\left(g\right)\to 8S{O}_2\left(g\right)$
$2S{O}_2\left(g\right)+O_2\left(g\right)\to 2S{O}_3\left(g\right)$
How many grams of $S{O}_3$ are produced from $1$ mole of $S_8$?

• A. $1280.0$
• B. $640.0$
• C. $960.0$
• D. $320.0$

Answer 1

Answer: (B)

$640.0$

${S_8}_{\left(s\right)}+{8O_2}_{\left(g\right)}⟶{8S{O}_2}_{\left(g\right)}$

${2S{O}_2}_{\left(g\right)}+{O_2}_{\left(g\right)}⟶{2S{O}_3}_{\left(g\right)}$

By stoichiometry, $1$ mole of ${S_8}_{\left(s\right)}$ produces $8$ mole ${S{O}_2}_{\left(g\right)}$

Also, $2$ mole of ${S{O}_2}_{\left(g\right)}$ produces $2$ mole ${S{O}_3}_{\left(g\right)}$

Therefore, $8$ mole ${S{O}_2}_{\left(g\right)}$ produces $8$ moles ${S{O}_3}_{\left(g\right)}$

i.e. $1$ mole ${S_8}_{\left(s\right)}$ produces $8$ moles ${S{O}_3}_{\left(g\right)}$

i.e. $8\times 80g$ of $S{O}_3$

i.e $640g$ of $S{O}_3$