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agasfer [191]
3 years ago
6

the combustion of octane is given in this reaction 2C8H18+2502-----16CO2+18H2 O, identify the limit reagent for this reaction wi

th respect of carbon dioxide given starting mass of 12.85g of octane and 7.46g of oxygen.
Chemistry
1 answer:
Archy [21]3 years ago
4 0

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

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