Question

In the above reaction - 2SO2 + O2 = 2SO3 (irreversible) for the formation of one ton of SO3, what would be the quantity of O2 required

Answer 1

Answer:

250kg

Explanation:

2 mole so2 requires 1mole of O2

= 2×64/1000kg - - - - --1×32/1000kg

now 1000kg (1ton) - - - - - ?O2

1000*32/128 = 250kg

Answer 2

The quantity of $O_2$ required is 0.2 ton

Explanation:

According to avogadro's law, 1 mole of every substance weighs equal to molecular mass and contains avogadro's number $6.023\times 10^{23}$ of particles.

To calculate the number of moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$

given mass of $SO_3$ = 1 ton = 1000 kg = $10^6g$

Molar mass of $SO_3$ = 80 g/mol

Putting in the values we get:

$\text{Number of moles}=\frac{10^6g}{80g/mol}=12500moles$

$2SO_2+O_2\rightarrow 2SO_3$

According to stoichiometry:

2 moles of $SO_3$ are produced from 1 mole of $O_2$

12500 moles of $SO_3$ are produced from =$\frac{1}{2}\times 12500=6250$ moles of $O_2$

Mass of oxygen=$\moles\times {\text {Molar mass}}=6250mol\times 32g/mol=2\times 10^5g=0.2 ton$    (1ton=1000kg) (1kg=1000g)

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