Question

calculate mass of so3 formed, if 500g so2 react with 200g o2. according to the equation identify the limting reagent​

Answer 1

Answer:

Answer: 625 grams and SO_2SO

2

is the limiting reagent .

Explanation:

To calculate the moles, we use the equation:

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

Molar mass

Given mass

a) moles of SO_2SO

2

\text{Number of moles}=\frac{500g}{64g/mol}=7.81molesNumber of moles=

64g/mol

500g

=7.81moles

b) moles of O_2O

2

\text{Number of moles}=\frac{200g}{32g/mol}=6.25molesNumber of moles=

32g/mol

200g

=6.25moles

2SO_2(g)+O_2(g)\rightarrow 2SO_3(g)2SO

2

(g)+O

2

(g)→2SO

3

(g)

According to stoichiometry :

2 moles of SO_2SO

2

require 1 mole of O_2O

2

Thus 7.81 moles of SO_2SO

2

require=\frac{1}{2}\times 7.81=3.91moles

2

1

×7.81=3.91moles of O_2O

2

Thus SO_2SO

2

is the limiting reagent as it limits the formation of product.

As 2 moles of SO_2SO

2

give = 2 moles of SO_3SO

3

Thus 7.81 moles of SO_2SO

2

give =\frac{2}{2}\times 7.81=7.81moles

2

2

×7.81=7.81moles of SO_3SO

3

Mass of SO_3=moles\times {\text {Molar mass}}=7.81moles\times 80g/mol=625gSO

3

=moles×Molar mass=7.81moles×80g/mol=625g

Thus mass of SO_3SO

3

produced is 625 grams and SO_2SO

2

is the limiting reagent .