Question

the combustion of octane is given in this reaction 2C8H18+2502-----16CO2+18H2 O, identify the limit reagent for this reaction with respect of carbon dioxide given starting mass of 12.85g of octane and 7.46g of oxygen.

Answer 1

Answer:

Liming reagent in the given reaction is oxygen.

Explanation:

Mass of octane = 12.85 g

Mass of oxygen = 7.46 g

Molar mass of octane = 114.23 g/mol

Molar mass of oxygen = 32

$Mole=\frac{Mass\;in\;g}{Molar\;mass}$

$No.\;of\;moles\;of\;octane=\frac{12.85}{114.23} = 0.1125\;mol$

$No.\;of\;moles\;of\;oxygen=\frac{7.46}{32} = 0.2331\;mol$

$2C_8H_{18} + 25O_2 \rightarrow 16CO_2 +18H_2O$

According to the reaction, 2 moles of octane requires 25 moles of O2

so, 0.1125 moles of octane requires $0.1125\times \frac{25}{2}  = 1.40625\;mol$ of oxygen.

but only 0.2331 mol of oxygen is present.

Hence, oxygen is the limiting reagent