Question

If 6 gm of Carbon Reacts with 6 gm of Oxygen to produce 21 it of
CO2. The percentage yield of Reaction
100
50
75
25​

Answer 1

Answer:

25

Explanation:

It is your answer hope it's help you.

Answer 2

Given info : 6 gm of Carbon reacts with 6 gm of Oxygen to produce 2.1 lit of CO₂.

To find : The percentage yield of reaction is...

solution : first we have to see how carbon and oxygen reacts with each other.

here, C + O₂ ⇒ CO₂

so 1 mol of carbon and 1 mole of oxygen react to produce 1 mole of carbon dioxide.

given, mass of carbon = 6 g

atomic mass of carbon = 12 g/mol

so no of moles of carbon = 6/12 = 0.5 mol

now mass of oxygen = 6 g

but molar mass of oxygen = 32 g/mol

∴ no of moles of oxygen = 6/32 = 0.1875 mol

you see, oxygen is very small in amount with respect to carbon. so oxygen here will be limiting reagent.

means, reaction will stop after completing the consumption of oxygen gas.

from above , 1 mole oxygen forms 1 mole carbon dioxide

so, 0.1875 mole of oxygen forms 0.1875 carbon dioxide.

AT STP , 1 mole of a compound = 22.4 L

so, 0.1875 mole of carbon dioxide = 0.1875 × 22.4 L = 4.2 L

but actual yield = 2.1 L

now percentage yield = actual yield/theoretical yield × 100

= 2.1/4.2 × 100 = 50 %