Question

How many grams of NaCl are produced when 80.0 grams of ${{O}_{2}}$ are produced?
$2NaCl{{O}_{3}}\to 2NaCl+3{{O}_{2}}$

Answer 1

Write the balanced chemical equation and then find out the ratio of the production of sodium chloride, then use stoichiometry to calculate the weight produced.

Complete answer:
In order to answer the question, we need to know about moles and molar mass. Now, matter is made up of atoms, and as matter has mass, then the atoms should have an individual mass. Molar mass of an element or compound is the mass which houses $6\times {{10}^{23}}$ particles. For, example, the hydrogen molecule has a molar mass of 2 grams. This means 2 grams of hydrogen contains $6\times {{10}^{23}}$atoms, and this number is also called the Avogadro’s number.
Number of moles of an element or a compound is the ratio of its given mass taken by the user, to its molar mass. More is the number of moles, more is the concentration of the substance. Now, let us come to the question. We first write the balanced chemical equation for the reaction an analyse:
$2NaCl{{O}_{3}}\to 2NaCl+3{{O}_{2}}$
2 moles of NaCl gives 3 moles of oxygen molecule, and the molar mass of NaCl and oxygen molecule are $58g\,mo{{l}^{-1}},32g\,mo{{l}^{-1}}$ respectively. So, now we use the mole ratio of sodium chloride and oxygen molecule to find the grams of sodium chloride produced. On calculating, we have:

& {{w}_{NaCl}}=80g{{O}_{2}}\times \left( \dfrac{1mol\,{{O}_{2}}}{32g\,{{O}_{2}}} \right)\times \left( \dfrac{2mol\,NaCl}{3mol\,{{O}_{2}}} \right)\times \left( \dfrac{58g\,NaCl}{1mol\,NaCl} \right) \\\ & \Rightarrow {{w}_{NaCl}}=97.5g \\\ \end{aligned}$$ So, if we take 80 grams of the oxygen molecule, then approximately 97.5 grams of sodium chloride will be produced. **Note:** The following method is used in order to calculate the mass to mole calculations: $$Quantity\,sought=Quantity\,given\times Conversion\,factor$$ Also, the number of moles is a dimensionless quantity.