Question

$ 6grams $ of carbon was completely burnt in oxygen. What would be the volume of $ C{O_2} $ produced at NTP and how many molecules will be present in that gas?

Answer 1

Hint: Chemical equations, when written in the balanced form can give quantitative relationships between various reactants and products in terms of moles, masses, molecules and volumes. The quantitative information conveyed by a chemical equation helps in a number of calculations.

Complete answer:
In the question, it is given that $ 6 $ grams of carbon was completely burnt in oxygen;
 $ C + {O_2} \to C{O_2} $
We have to find the volume of carbon dioxide produced at NTP and the number of molecules present in the gas.
Molar mass of carbon= $ 12gmo{l^{ - 1}} $
Molar mass of Carbon dioxide $ \left( {C{O_2}} \right) $ = $ 44gmo{l^{ - 1}} $
According to the balanced chemical equation;
 $ 12g $ of carbon completely burns to give $ 44g $ of carbon dioxide $ \left( {C{O_2}} \right) $
We can find the amount of $ C{O_2} $ produced with $ 6g $ of carbon;
 $ 12g $ of carbon $ \to $ $ 44g $ of $ C{O_2} $
 $ 6g $ of carbon $ \to $ $ \dfrac{{44 \times 6}}{{12}}g $
 $ \to 22g $ of $ C{O_2} $
We can find the moles of $ C{O_2} $ produced;
 $ Moles = \dfrac{{mass}}{{mol.mass}} $
 $ = \dfrac{{22}}{{44}} $
 $ = 0.5 $ moles
At NTP, the volume occupied by one mole of gas is $ 22.4L $
 $ 1 $ mole of $ C{O_2} $ $ \to $ $ 22.4L $
 $ 0.5 $ mole of $ C{O_2} $ $ \to $ $ 22.4 \times 0.5 $
 $ \to $ $ 11.2L $
Thus, $ 11.2 L $ of carbon dioxide ( $ C{O_2} $ ) was produced.
Now, we have to find the number of molecules present in the gas.
We know that;
One mole of a gas contains Avogadro number ( $ 6.022 \times {10^{23}} $ ) of molecules.
We can find the number of molecules present in $ 0.5 $ mole of gas;
 $ 1 $ mole contain $ \to $ $ 6.022 \times {10^{23}} $ molecules
 $ 0.5 $ mole contain $ \to $ $ 6.022 \times {10^{23}} \times 0.5 $
 $ \to $ $ 3.011 \times {10^{23}} $ molecules
Thus, the volume of carbon dioxide was found to be $ 11.2L $ and the number of molecules was found to be $ 3.011 \times {10^{23}} $ .

Note:
The molar masses should be known and one should also know about the NTP conditions as well in order to solve such questions. The equations should be balanced before equating the stoichiometry.