Question

Gaseous butane CH3CH22CH3 will react with gaseous oxygen O2 to produce gaseous carbon dioxide CO2 and gaseous water H2O. Suppose 5.81 g of butane is mixed with 35. g of oxygen. Calculate the minimum mass of butane that could be left over by the chemical reaction. Be sure your answer has the correct number of significant digits.

Answer 1

Amount of butane = 5.81 g

Molar mass of butane = 58.1 g/mol

So, no. of Mole of butane = (5.81/58.1) mole

= 0.1 mole

Amount of O2 = 35 g of oxygen

Molar mass of oxygen = 32 g/mol

So, no. of Mole of oxygen = 1.094 mol

Gaseous butane CH3(CH2)2CH3 reacts with O2 according to the following equation,

2CH3(CH2)2CH3 + 13O2 ---> 8CO2 + 10H2O

From the above balanced chemical equation,

2 moles butane react with 13 mole of O2

1 mole butane react with (13/2) mole O2

0.1 Mole of butane reacts with

(13×0.1/2) moles of butane

= 0.65 mol oxygen

But here 1.094 mole of oxygen is supplied.

so, no of Mole of oxygen left

= (1.094–0.65) mol

= 0.444 mol

0.444 mol oxygen = (32×0.444)g

= 14.208 g

So minimum 14.208 g of oxygen will be left over. No more butane remain.