Question

Se combinan 40g de SO2 y 25g de O2, determine el porcentaje en masa del exceso con respecto a su masa inicial. So2 + O2 --> SO3
P.A.(S=32 ; O=16)

Answer 1

Answer : The percentage mass of excess with respect to initial mass is 60.0 %

Solution : Given,

Mass of $SO_2$ = 40 g

Mass of $O_2$ = 25 g

Molar mass of $SO_2$ = 64 g/mole

Molar mass of $O_2$ = 32 g/mole

First we have to calculate the moles of $SO_2$ and $O_2$.

$\text{ Moles of }SO_2=\frac{\text{ Mass of }SO_2}{\text{ Molar mass of }SO_2}=\frac{40g}{64g/mole}=0.625mole$

$\text{ Moles of }O_2=\frac{\text{ Mass of }O_2}{\text{ Molar mass of }O_2}=\frac{25g}{32g/mole}=0.781mole$

Now we have to calculate the limiting and excess reagent.

The balanced chemical reaction is,

$2SO_2(s)+O_2(g)\rightarrow 2SO_3(s)$

From the balanced reaction we conclude that

As, 2 mole of $SO_2$ react with 1 mole of $O_2$

So, 0.625 moles of $SO_2$ react with $\frac{0.625}{2}=0.312$ moles of $O_2$

From this we conclude that, $O_2$ is an excess reagent because the given moles are greater than the required moles and $SO_2$ is a limiting reagent and it limits the formation of product.

Now we have to calculate the excess moles of $O_2$

Excess moles of $O_2$ = Required moles - Given moles

Excess moles of $O_2$ = 0.781 - 0.312 = 0.469 mole

Now we have to calculate the excess mass of $O_2$

$\text{ Mass of }O_2=\text{ Moles of }O_2\times \text{ Molar mass of }O_2$

$\text{ Mass of }O_2=(0.469moles)\times (32g/mole)=15.0g$

Excess mass of $O_2$ = 15.0 g

Initial mass of $O_2$ = 25 g

Now we have to calculate the percentage mass of excess with respect to initial mass.

$\text{Percentage}=\frac{\text{ Excess mass of }O_2}{\text{ Initial mass of }O_2}\times 100$

$\text{Percentage}=\frac{15.0g}{25g}\times 100=60.0\%$

Therefore, the percentage mass of excess with respect to initial mass is 60.0 %