Question

How many grams of Ca does 32 grams of ${{O}_{2}}$ (oxygen) have to react to give 72 grams of CaO? Answer is given 40 grams but if 32 grams of oxygen has to react then it means that there are two oxygen atoms and then the formula will be $Ca{{O}_{2}}$?

Answer 1

Hint: The reaction of calcium and oxygen produces calcium oxide also called quicklime. The law of conservation of mass states that mass of reactants is equal to mass of products.

Complete answer: We have been given to find the amount of Calcium (in grams), when 32 g of oxygen produces 72 g of calcium oxide. It is also given that 40 g of calcium, but that would make calcium dioxide as the product. For this the reaction is written as,
$Ca(s)+\dfrac{1}{2}{{O}_{2}}(g)\to CaO(g)$
As we can see, 1 mole of Ca produces 1 mole of CaO. So, we have to calculate the number of moles of CaO in 72 grams of CaO as,
Number of moles of CaO in 72 g = $\dfrac{given\,mass\,of\,CaO}{molar\,mass\,of\,CaO}$
Number of moles of CaO in 72 g =$\dfrac{72\,g}{56.077\,g\,mo{{l}^{-1}}}$
Number of moles of CaO in 72 g = 1.2839 moles
Now we will obtain the number of moles of calcium in 1.2839 moles of CaO through stoichiometric calculations as,
Moles of Ca in 1.2839 moles of CaO = 1.2839 mol CaO $\times \dfrac{1\,mol\,Ca}{1\,mol\,CaO}$
Moles of Ca in 1.2839 moles of CaO = 1.2839 mol Ca
Now we will calculate the mass in these moles of calcium using the relationship between moles and mass as,
Number of moles = $\dfrac{mass}{molar\,mass}$
Number of moles of Ca = $\dfrac{mass}{molar\,mass\,of\,Ca}$
1.2839 mol Ca = $\dfrac{mass}{40.07\,g}$
Therefore, mass = 1.2839 mol$\times $ 40.07 g
Mass of Ca = 51.4 g
Hence, 51.4 gram of calcium is needed to produce 72 gram of CaO, also oxygen is given in excess. This is the reason 32 grams of oxygen gives 1 mole of CaO and two oxygen atoms are not involved in the product.

Note: In the given data, oxygen is in excess and calcium is the limiting reagent. As by law of conservation of mass, only 20.6 g of oxygen is needed to yield 72 grams of CaO.