Question

12.8 g of so2 reacts with 3.2 g of O2 to form so3. if 6g of so3 is produced then calculate %yield of reaction.​

Answer 1

Explanation:

2SO

2

+O

2

→SO

3

Here the mole ratio between S0

2

and SO

3

is 2:2 = 1.

Here we used 2 g of SO

2

, so we need to convert that amount into moles.

Sulphur dioxide has molar mass of 64 g/mol. So,

64g/mol

2g

= 0.03125 mol of SO

2

Sulphur trioxide has molar mass of 80 g/mol. So,

0.03125 mol. 80 g/mol = 2.5 g of SO

3

Sp, 2.5 grams of sulphur trioxide was produced.

Answer 2

%yield of reaction is 100%.

Calculation:

Chemical reaction:

SO2 + O2 → SO3

Balanced chemical reaction:

2SO2 + O2 → 2SO3

Hence,

2 moles of SO2 will react with 1 mole of O2 to form 2 moles of SO3 .

So, Molar mass of 1 mole of SO2 = 32+ 2( 16) = 64 g

mass of 2 moles of SO2 = 64+ 64 = 128 g

1 mole of O 2 Is used , molar mass of 1 mole of O2 = 32 g

Molar mass of 1 mole of SO3= 32+ 3(16) = 80 g

mass of 2 moles of SO3 = 160 g

Hence in terms of mass balance we can write,

64g SO2 + 32 g O2 → 80 g SO3

But in the given question..

12.8 g SO2 + 3.2 g O2 → ? g SO3

Here, we can say that 2 moles of SO2 will give 2 moles of 2 moles of SO3 ..

So we can say that..

If , 128 g SO2 will give 160 g SO3

so , 12.8 g SO2 will give how many g of SO3?

= 12.8× 160 ÷ 128

=16 g SO3 will produce in the given reaction.

moles of SO2 used = 12.8/ 64 = 0.2

Hence, 0.1 moles of O2 should needed for this reaction..

Here, given 3.2 g of O2 .

moles of O2 = mass / 32

0.1 × 32 = mass

Hence, all the O2 will get consumed during the reaction.

Total mass of the Reactants= 12.8 + 3.2 = 16 g

% yield of the reaction= mass of product ×100 ÷ total mass of reactants

= 16 ×100 ÷ 16

= 100%

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