Question

Consider the following reaction sequence.
${S_{8(s)}} + 8{O_{2(g)}} \to 8S{O_{2(g)}}$
$2S{O_{2(g)}} + {O_{2(g)}} \to 2S{O_3}$
How many grams of $S{O_3}$ are produced from $1\,mole\,of\,{S_8}$?
A. $1280\,g$
B. $960\,g$
C. $640\,g$
D. $320\,g$

Answer 1

Hint:In order to the solution, first we should find the number of moles contained by $S{O_3}$compound in both of the reactions. Now, we have both the number of moles in $S{O_3}$ and its molar mass. Then, we can find its weight from here.

Complete answer:
In first reaction:
${S_{8(s)}} + 8{O_{2(g)}} \to 8S{O_{2(g)}}$
As we have $1\,mole\,of\,{S_8}$.
According to reaction, one mole of $S{O_2}$ produces 8 mole of $S{O_2}$.
Further, in second reaction:-
 $2S{O_{2(g)}} + {O_{2(g)}} \to 2S{O_3}$
As per the reaction
Number of moles of$S{O_3}$, produced by $2\,moles\,of\,S{O_2} = 2\,moles$
$\therefore $ Number of moles of $S{O_3}$, produced by $8\,moles\,of\,S{O_2} = 8\,moles$.
As we know that,
Finally, we know the both number of moles and the molar mass of $S{O_3}$ in $8\,moles$, So we can find the weight of this compound by the formula:
Weight of compound = $no.\,of\,moles \times molar\,mass$
$\therefore $ Weight of $S{O_3}$ in $ 8\,moles = 8 \times 80 = 640 g [\ because\, Molar\,Mass\,of\,S{O_2} = 80\,g]$
Hence, $640\,gm\,of\,S{O_3}$ are produced by $1\,mole\,of\,{S_8}$.

So, the correct option is (C). $640\,g$.

Note:
Molar mass of any compound can be determined by adding the standard nuclear masses of the constituent particles. For instance: the molar mass of or weight of water is 18.015 nuclear mass units, so one mole of water weight .