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Question: Sulphur trioxide is prepared by the following two reactions: \[{{S}_{8(s)}}+8{{O}_{2(g)}}\to 8S{{O...

Sulphur trioxide is prepared by the following two reactions:
S8(s)+8O2(g)→8SO2(g){{S}_{8(s)}}+8{{O}_{2(g)}}\to 8S{{O}_{2(g)}} and 2SO2(g)+O2(g)→2SO3(g)2S{{O}_{2(g)}}+{{O}_{2(g)}}\to 2S{{O}_{3(g)}}
How many grams of SO3(g)S{{O}_{3(g)}}​ are produced from one mole of S8(s){{S}_{8(s)}} ​?
A. 12801280
B. 640640
C. 960960
D. 320320

Explanation

Solution

As we have already learnt that a mole is the definite amount of a substance expressed in terms of weight, volume and number of particles. One mole contains a fixed number of particles as present in 12g12g of Carbon−12Carbon-12 atoms.

Complete step by step answer:
So, to calculate the number of molecules, we can first identify the moles and then using the formula given below we can calculate the molecules without any difficulties. Moles can be easily calculated by dividing the given mass of the substance with the molecular mass of the given substance. Then, with the calculated moles, the number of molecules can be calculated by multiplying the moles with Avogadro’s number. For that let us first write the values given in the question,
Sulphur trioxide is prepared by the following two reactions:
S8(s)+8O2(g)→8SO2(g){{S}_{8(s)}}+8{{O}_{2(g)}}\to 8S{{O}_{2(g)}} and 2SO2(g)+O2(g)→2SO3(g)2S{{O}_{2(g)}}+{{O}_{2(g)}}\to 2S{{O}_{3(g)}}
By stoichiometry, one mole of S8(s){{S}_{8(s)}} produces 88 mole SO2(g)S{{O}_{2(g)}} Also, two mole of SO2(g)S{{O}_{2(g)}} produces two mole SO3(g)S{{O}_{3(g)}}
Similarly, the number of atoms can be calculated by first calculating the moles and then multiplying those calculated moles with Avogadro’s number. If volume is given and we are asked to calculate the number of atoms or number of molecules then we can apply the formula;
Therefore, 88 mole SO2(g)S{{O}_{2(g)}}​ produces 88 moles SO3(g)S{{O}_{3(g)}}
i.e. One mole S8(s){{S}_{8(s)}}produces 88 moles
⇒8×80g[SO3(g)]\Rightarrow 8\times 80g\left[ S{{O}_{3(g)}} \right]
⇒640g[SO3(g)]\Rightarrow 640g\left[ S{{O}_{3(g)}} \right]

So, the correct answer is Option B.

Note: Students should learn this formula to calculate anything given in the question so that you do not get any answer wrong.
moles=massmolecular(mass)=no.(molecules)6.022×1023=no.(atoms)6.022×1023=given(volume)22.4Lmoles=\dfrac{mass}{molecular(mass)}=\dfrac{no.(molecules)}{6.022\times {{10}^{23}}}=\dfrac{no.(atoms)}{6.022\times {{10}^{23}}}=\dfrac{given(volume)}{22.4L}
This is the easiest formula to learn and from an entrance exams point of view it takes less time to calculate the answers.