Question

10g of CO on burning in air gives the product which have the number of nucleons​

Answer 1

Answer:

Hydrogen + Oxygen → Carbon dioxide + Water

↓ ↓

10 gm 40 g

As when CO

2

is absorbed in lime water, the mass of soln increases by 27.5 gm

∴ Mass of CO

2

=27.5 gm

∴ Mass of H

2

O= (Mass of hydrogen + Mass of oxygen − mass of CO

2 )

=(10+40−27.5) gm

=22.5 gm

Answer 2

Given:

Mass of $CO=10g$

To Find: Number of nucleons in the product produced on burning $10 g$ of  $CO$ in air

Solution:

The reaction of burning of carbon dioxide in the air can be given as:

$\[2CO + {O_2} \to 2C{O_2}\]$

The number of moles present in 10 g  of CO can be calculated as:

$\[No.\,of\,moles = \frac{{Mass}}{{Molar\,mass}}\]$

$\[No.\,of\,moles = \frac{{10g}}{{28}}\]$

$\[No.\,of\,moles = 0.357\]$

According to the above reaction,

2 moles of CO produces 2 moles of $\[C{O_2}\]$

Therefore,

0.357 moles of CO will produce 0.357 moles of $\[C{O_2}\]$

Number of nucleons in 1 mole of $\[C{O_2} = 12 + 2 \times 16 = 44\]$

Thus, the number of nucleons in 0.357 moles of $\[C{O_2}\]$ can be given as:

$\[No.\,of\,nucleons = Nucleons\,in\,1\,molC{O_2} \times moles\,of\,C{O_2} \times Avogadro\,number\]$

$\[No.\,of\,nucleons = 44 \times 0.357 \times 6.023 \times {10^{23}}\]$

$\[No.\,of\,nucleons = 94.6 \times {10^{23}}\]$

Hence, the number of nucleons in the product produced on burning 10g of CO in air is $\[94.6 \times {10^{23}}\]$.