One mole of butane is burnt in excess oxygen and the gas produced is allowed to pass through a 3 M lime water solution having 1 L volume producing $C a C O_{3}$ and $H_{2} O$ . The mole(s) of $C a C O_{3}$ produced are .

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Solution

Combustion of hydrocarbons produces $C O_{2}$ gas.

$C_{4} H_{10} + \frac{13}{2} O_{2} \to 4 C O_{2} + 5 H_{2} O$
1 mol of butane produces 4 mol of $C O_{2}$ .
Now $C O_{2}$ produced is allowed to pass through $C a (O H)_{2}$ , the reaction is:
$C O_{2} + C a (O H)_{2} \to C a C O_{3} + H_{2} O$
$4 m o l 3 m o l ?$
Here the concept of limiting reagent comes into the picture and turns out that $C a (O H)_{2}$ is the limiting reagent here.

Since only $3 m o l$ of $C a (O H)_{2}$ is present, hence $3 m o l$ of $C a C O_{3}$ will be formed.

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