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

Liming reagent in the given reaction is oxygen.

Step-by-step 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=(Mass\;in\;g)/(Molar\;mass)`

`No.\;of\;moles\;of\;octane=(12.85)/(114.23) = 0.1125\;mol`

`No.\;of\;moles\;of\;oxygen=(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* (25)/(2)  = 1.40625\;mol` of oxygen.

but only 0.2331 mol of oxygen is present.

Hence, oxygen is the limiting reagent