The reaction of burning of carbon in oxygen is represented by equation
$C_{(s)} + O_{2 (g)} \to C O_{2 (g)} + h e a t + L i g h t$

When 9.0 g of solid carbon in burnt in 16.0 g of oxygen gas the mass of carbon dioxide gas formed would be :

(Note: atomic mass of $C = 12.0 μ, O = 16.0 μ$ )

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25.0 g

2.33 g

22.0 g

33.00 g

answer is B.

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Detailed Solution

i.e. 22.0 g of

C O_{2}

is formed. During this reaction, we have the reaction

$C_{(s)} + O_{2 (g)} \to C O_{2} + H e a t + L i g h t$

According to mole concept :

1 mole of carbon + 1 mole of $O_{2} \to 1 m o l e o f C O_{2}$

i.e. 12g of carbon $+ 32$ gm of oxygen $\to$ 44 gm of $C O_{2}$ Now we have 9.0 g of carbon & 16.0 g of $O_{2}$ in this reaction and 9.0 g of carbon will require 24 g of $O_{2}$

to react with it completely. Therefore, in this reaction, $O_{2}$ gas is limiting reagent and the amount of $C O_{2}$ produced will depend upon the quantity of $O_{2 (g)}$ present in the reaction. Therefore, we can calculate the amount of $C O_{2}$ gas as 32 gm of $O_{2}$ produces : 44 gm of $C O_{2}$ 16 gm of $O_{2}$ will produce : 22 gm of $C O_{2}$ .