Question

(сн, сH), сн-) (CH will react with gaseous oxygen (0, to produce gaseous carbon dioxide CO2 and gaseous water (H,O). Suppose Gaseous butane 2.3 q of butane is mixed with 11.6 q of oxyqen. Calculate the minimum mass of butane that could be left over by the chemical reaction. Round your answer to 2 significant digits. ? X

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)= 2.3 g

use:

number of mol of C4H10,

n = mass of C4H10/molar mass of C4H10

=(2.3 g)/(58.12 g/mol)

= 3.957*10^-2 mol

Molar mass of O2 = 32 g/mol

mass(O2)= 11.6 g

use:

number of mol of O2,

n = mass of O2/molar mass of O2

=(11.6 g)/(32 g/mol)

= 0.3625 mol

Balanced chemical equation is:

2 C4H10 + 13 O2 ---> 8 CO2 + 10 H2O

2 mol of C4H10 reacts with 13 mol of O2

for 3.957*10^-2 mol of C4H10, 0.2572 mol of O2 is required

But we have 0.3625 mol of O2

so, C4H10 is limiting reagent

Since CH10 is limiting reagent, there should be no butane left over.

Answer: 0.0 g