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Limiting Reagent or Reactant

Consider the ...

Question

Consider the following combustion reaction:
$C_{3} H_{8} (g) + 5 O_{2} (g) \to 3 C O_{2} (g) + 4 H_{2} O (g)$
At STP (atm), when 64.4 L of propane is reacted with 87.0 L of oxygen gas, volume of the $C O_{2}$ gas produced is:

105 L of

C O_{2}

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258 L of

C O_{2}

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39.2 L of

C O_{2}

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52.2 L of

C O_{2}

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Solution

The correct option is D 52.2 L of $C O_{2}$
1 mole of propane requires 5 moles of oxygen
22.4 L of propane requires $(5 \times 22.4)$ L of oxygen
64.4 L of propane will require $\frac{5 \times 22.4}{22.4} \times 64.4 = 322$ L of oxygen
Therefore, oxygen is the limiting reagent.
5 moles of oxygen produce 3 moles of $C O_{2}$
$(5 \times 22.4)$ L of oxygen produces $(3 \times 22.4)$ L of $C O_{2}$
So, 87 L of oxygen will produce $\frac{3 \times 22.4}{5 \times 22.4} \times 87$ = 52.2 L of $C O_{2}$

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Limiting Reagent or Reactant

Standard XII Chemistry

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Consider the following combustion reaction: C3H8(g)+5O2(g)→3CO2(g)+4H2O(g) At STP (atm), when 64.4 L of propane is reacted with 87.0 L of oxygen gas, volume of the CO2 gas produced is:

Consider the following combustion reaction:
$C_{3} H_{8} (g) + 5 O_{2} (g) \to 3 C O_{2} (g) + 4 H_{2} O (g)$
At STP (atm), when 64.4 L of propane is reacted with 87.0 L of oxygen gas, volume of the $C O_{2}$ gas produced is: