Question

68g of methane reacts with 80g of o2 to form co2 and H2O identify limiting reagent and calculate no of moles​

Answer 1

Given info : 68g of methane reacts with 80g of O2 to form CO2 and H2O.

To find : limiting reagent and also no of moles of products.

solution : given weight of methane = 68

molecular weight of methane = 16 g/mol

so, no of moles of methane = 68/16 = 4.25

Similarly, no of moles of oxygen gas = given weight/molecular weight

= 80/32 = 2.5

reaction of methane with oxygen gas is given by,

CH₄ + 2O₂ ⇒CO₂ + 2H₂O

From above it is clear that, 1 mole of methane reacts with 2 moles of oxygen.

so, 4.25 moles of methane react with 8.5 moles of oxygen.

But here no of moles of oxygen is 2.5

so, oxygen is limiting reagent.

Because 2 moles of oxygen produce one mole of CO₂ and two moles of water.

So, 2.5 moles of oxygen produce 1.25 moles of CO₂ and 2.5 moles of water.

Therefore the no of moles of carbon dioxide = 1.25

no of moles of water = 2.5

Answer 2

Answer:-

Given info :

68g of methane reacts with 80g of O2 to form CO2 and H2O.

To find :

limiting reagent and also no of moles of products.

solution :

given weight of methane = 68

molecular weight of methane = 16 g/mol

so, no of moles of methane = 68/16 = 4.25

Similarly, no of moles of oxygen gas = given weight/molecular weight

= 80/32 = 2.5

reaction of methane with oxygen gas is given by,

CH₄ + 2O₂ ⇒CO₂ + 2H₂O

From above it is clear that, 1 mole of methane reacts with 2 moles of oxygen.

so, 4.25 moles of methane react with 8.5 moles of oxygen.

But here no of moles of oxygen is 2.5

so, oxygen is limiting reagent.

Because 2 moles of oxygen produce one mole of CO₂ and two moles of water.

So, 2.5 moles of oxygen produce 1.25 moles of CO₂ and 2.5 moles of water.

Therefore the no of moles of carbon dioxide = 1.25

no of moles of water = 2.5

Hope it's help you❤️