Calculate the amount of Carbon monoxide formed when 840 g of hexene reacts with 1 kg of oxygen. (Assume no $C O_{2}$ is formed).

27.54 mol

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18.96 mol

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13.65 mol

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31.25 mol

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Solution

The correct option is D 31.25 mol
Balanced reaction : $C_{6} H_{12} + 6 O_{2} \to 6 C O + 6 H_{2} O$
We know that,
Number of moles = $\frac{G i v e n m a s s}{M o l a r m a s s}$

Number of moles of $C_{6} H_{12}$ = $\frac{840}{84}$ = 10 mol
Number of moles of $O_{2}$ = $\frac{1000}{32}$ = 31.25 mol
We know that whenever moles of more than one reactant is given, we need to determine the Limiting Reagent (LR) to find the product of the reaction.
Calculating LR,
For Hexene = $\frac{n u m b e r o f m o l e s}{s t o i c h i o m e t r i c c o e f f i c i e n t}$ = $\frac{10}{1}$ = 10
For Oxygen = $\frac{n u m b e r o f m o l e s}{s t o i c h i o m e t r i c c o e f f i c i e n t}$ = $\frac{31.25}{6}$ = 5.20

Since the above ratio is less for Oxygen, the limiting reagent will be Oxygen.
As per stoichiometry, 6 mol of $O_{2}$ reacts to form 6 mol of $C O$
So, 31.25 mol will form 31.25 mol of $C O$

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