Question

Consider the reaction,
2C4H10(g) + 13O2(g) --> 8CO2(g)+10H2O(l)
If 300g of C4H10
(g) is burnt with 1000 g of O2

, the yield of H2O and the limiting reagent in the reaction

are _____ and _____ respectively.

a. 435 g and O2

b. 400 g and O2

c. 350 g and C4 H10

d. 465 g and O2​

Answer 1

Given: 2C4H10(g) + 13O2(g) --> 8CO2(g)+10H2O(l)

300g of C4H10 (g) is burnt with 1000 g of O2.

To find: Yield of H20 and limiting reagent in the reaction

Explanation: 2 moles of butane(C4H10) reacts with 13 moles of O2 to produce 8 moles of CO2 and 10 moles of H2O.

Now, given mass of C4H10= 300 g

Molar mass of C4H10= 58 g

Moles of C4H10

= Given mass/ Molar mass

= 300/58

= 5.17 moles

Now, given mass of O2= 1000 g

Molar mass of C4H10= 32 g

Moles of O2

= Given mass/ Molar mass

= 1000/32

= 31.25 moles

Now, we see in the question that 2 mole of C4H10 requires 13 moles of O2 for complete reaction.

1 mole of C4H10 will require 13/2 moles of O2

Therefore, 5.17 moles of C4H10 will require

= 13/2 * 5.17

= 33.06 moles of O2 for complete reaction but only 31.25 moles are present.

This means that O2 is not sufficient for complete reaction. It is the limiting reagent.

Now, 13 moles of O2 gives 10 moles H20

1 moles of O2 would give 10/13 moles H20

Then, 31.25 moles of O2 will give

= 10/13 * 31.25

= 24.1 moles of H2O

Therefore, mass of H20

= Moles produced* Molar mass of H20

= 24.1 * 18

= 435 g

O2 is the limiting reagent and 435 g of H2O is produced. Therefore, option (a) is the correct answer

Answer 2

Answer:

Explanation:

A