Question

The reaction of burning of carbon in oxygen is represented by the equation:
$C_{\left(s\right)}+O_{2\left(g\right)}\to {CO}_{2\left(g\right)}+Heat+Light$

When 9.0 g of solid carbon is burnt in 16.0 g of oxygen gas, 22.0 g of carbon dioxide is produced. The mass of carbon dioxide gas formed on burning of 3.0 g of carbon in 32.0 g of oxygen would be (Note : atomic mass of $C=12.0u,O=16.0u$ ):

• A. 6.60 g
• B. 7.33 g
• C. 8.25 g
• D. 11.00 g

Answer 1

Answer: (D)

11.00 g

$C\left(s\right)+O_2\left(g\right)\to C{O}_2\left(g\right)$

9 gm of $C\left(s\right)=$0.75 mole of C(s)
16 g of $O_2\left(g\right)$ = 0.5 mole of $O_2\left(g\right)$
We know from the reaction that equal moles of C and $O_2$ burn and form equal mole of $C{O}_2$
According to the question, 0.5 mole each of carbon and oxygen form 0.5 mole of $C{O}_2$
In the above process 0.25 mole (3 g) of Carbon left out. Left amount of carbon reacts with 32.0 g (1 mole ) of $O_2$ to form 0.25 mole of $C{O}_2$.
Hence the amount of $C{O}_2$ form $=$ no. of moles $\times$ molecular mass of $C{O}_2$

$=0.25\times 44=11.0g$
Option D is correct.