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Atomic Number and Mass Number

The drug nico...

Question

The drug nicotine has the molecular formula $C_{10} H_{14} N_{2}$ . If 0.1 mole of this is combusted, what would be the weight of $C O_{2}$ obtained?

440 g

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4.4g

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44g

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100g

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Solution

The correct option is C 44g
$C_{10} H_{14} N_{2} + 27 / 2 O_{2} \to 10 C O_{2} + 7 H_{2} O + N_{2}$
1 mole 10 moles
0.1 mole 1 mole
Weight of $C O_{2}$ obtained $=$ no. of moles x M $=$ 1 x 44 $=$ 44g

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