Question

5.8 Kg of Butane is burned in 99.6 L of O2 at N.T.P. Identify (a) the limiting reagent , (b) amount of Co2 formed , (c) amount of reagent left unreacted.​

Answer 1

Mass of Butane taken = 5.8 kg =5800 gram

Molar mass of Butane ($C_4H_{10}$):-

$=4*12+10\\ \\ \\48+10\\ \\ \\=58$

Number of moles of Butane will be :-

$=\dfrac{5800}{58}\\ \\ \\=100\:Moles$

$\rule{200}{2}$

Volume of Oxygen taken =99.6 L

Moles of Oxygen taken will be :-

$=\dfrac{99.6}{22.4}\\ \\ \\=4.45\:Moles$

$\rule{200}{2}$

$2C_4H_{10}+13O_2 ---->8CO_2+10H_2O$

$\rule{200}{2}$

Clearly,

Oxygen is the Limiting Reagent Here:-

$\rule{200}{2}$

13 mole of Oxygen reacts to form 8 moles of CO2

4.45 moles of Oxygen forms:-

$=\dfrac{8}{13}*4.45\:Moles\:of\:CO_2\\ \\ \\=2.74\:Moles$

Amount of CO2 will be :-

$=2.74*44\\ \\ \\=120.56\:Grams$

$\rule{200}{2}$

13 moles of Oxygen requires 2 moles of Butane

4.45 moles of Oxygen will require:-

$=\dfrac{2}{13}*4.45\\ \\ \\=0.68\:Moles$

Reagent Left :-

$=100-0.68\\ \\ \\=99.32$

Mass of Reagent Left :-

$=99.32*58\\ \\ \\=5760.5\:Grams$

$\rule{200}{2}$