Question

The amount of CO2 produced when 45g of ethane is treated with 3.5 mole of O2 ( 2C2H6+7 O2+=4CO2+6H2O )​

Answer 1

Answer:

In the equation

( 2C2H6+7 O2+=4CO2+6H2O )

we see that

2moles of C2H6 + 7 moles of  O2 ⇒ 4 moles of CO2

then if 45g of ethane (C2H6) reacts with 3.5 g of O2 then amount of O2 will be

45g of ethane = 1.5 moles of ethane

so we need to find

1.5 moles of ethane (C2H6) + 3.5 moles of O2 ⇒ x moles of CO2(here O2 is the limiting reagent as its completely used ) so

7 moles of O2 ⇒ 4 moles of CO2

3.5 moles of O2 ⇒ x moles of CO2

cross multiply , you'll get x=4× 3.5 /7

so x=2

x moles = 2 moles

2moles of CO2 molecular weight will be 2×44 = 88g

option B is correct.

Explanation:

limiting reagent can be found by this method;

according to the question

2 moles of ethane reacts with 7 moles of O2 , then

1.5 moles of ethane will react with how many moles of O2?

this can be solve  by cross multiplication and u get x=7×1.5/2=5.2

but only 3.5g moles of O2 is available so its completely used and some amount of ethane is left ,so  as we know that the left over reagent is excess reagent and completely used reagent is limiting reagent , there fore ethane is excess reagent and  O2 is limiting reagent.