Question

Gaseous butane

CH3CH22CH3

reacts with gaseous oxygen gas

O2

to produce gaseous carbon dioxide

CO2

and gaseous water

H2O

. If

36.4g

of water is produced from the reaction of

49.99g

of butane and

339.9g

of oxygen gas, calculate the percent yield of water.

Round your answer to

3

significant figures.

Answer 1

Molar mass of C4H10,
MM = 4*MM(C) + 10*MM(H)
= 4*12.01 + 10*1.008
= 58.12 g/mol


mass(C4H10)= 49.99

use:
number of mol of C4H10,
n = mass of C4H10/molar mass of C4H10
=(49.99 g)/(58.12 g/mol)
= 0.8601 mol

Molar mass of O2 = 32 g/mol


mass(O2)= 339.9

use:
number of mol of O2,
n = mass of O2/molar mass of O2
=(339.9 g)/(32 g/mol)
= 10.621 mol

Balanced chemical equation is:
2 C4H10 + 13 O2 ---> 10 H2O + 8 CO2


2 mol of C4H10 reacts with 13 mol of O2
for 0.2994 mol of C4H10, 1.946 mol of O2 is required
But we have 10.621 mol of O2

so, O2 is limiting reagent
we will use O2 in further calculation


Molar mass of H2O,
MM = 2*MM(H) + 1*MM(O)
= 2*1.008 + 1*16.0
= 18.016 g/mol

According to balanced equation
mol of H2O formed = (10/13)* moles of O2
= (10/13)*10.621
= 8.17 mol


use:
mass of H2O = number of mol * molar mass
= 8.17*18.02
= 147.223 g

% yield = actual mass*100/theoretical mass
= 36.4*100/147.223
= 24.724 %
Answer: 24.724  %