Question

How would you calculate the maximum number of grams of $ CaC{l_2} $ that could be produced from $ 74g $ of $ Ca{(OH)_2} $ in the following reaction: $ 2HCl + Ca{(OH)_2} \to 2{H_2}O + CaC{l_2} $ ?

Answer 1

Hint: Calculate the number of moles from $ 74g $ of $ Ca{(OH)_2} $ . Then, find the mole-to-mole ratio of $ Ca{(OH)_2}:CaC{l_2} $ and thus find the number of moles of $ CaC{l_2} $ produced in the reaction. Simply change the number of moles of $ CaC{l_2} $ to grams by multiplying it with the molar mass of $ CaC{l_2} $ .

Complete step by step solution:
Since the chemical equation is already balanced, we will simply proceed to calculate the number of moles of $ Ca{(OH)_2} $ .
Given mass of $ Ca{(OH)_2} = 74g $
To find the molar mass of $ Ca{(OH)_2} $ we will simply add the individual mass of each element present in the molecule.
Molar mass of $ Ca{(OH)_2} = $ mass of $ Ca + $ $ 2 \times $ mass of $ (O + H) $
 $ = 40 + 2 \times (16 + 1) $
We multiplied the mass of $ O $ and $ H $ because in the molecule of $ Ca{(OH)_2} $ there are two anions of $ O{H^ - } $ .
 $ = 40 + 34 = 74g $
Number of moles $ (n) $ of $ Ca{(OH)_2} = \dfrac{{given{\text{ }}mass}}{{molar{\text{ }}mass}} $
 $ n = \dfrac{{74}}{{74}} = 1 $ mol
Here, $ n \to $ number of moles of $ Ca{(OH)_2} $
Therefore, as per the given equation,
 $ 2HCl + Ca{(OH)_2} \to 2{H_2}O + CaC{l_2} $
 $ 1 $ mol of $ Ca{(OH)_2} $ produces $ 1 $ mol of $ CaC{l_2} $ when a sufficient amount of hydrochloric acid is added to $ Ca{(OH)_2} $ . Hence, the mole-to-mole ratio of $ Ca{(OH)_2}:CaC{l_2} $ is $ 1:1 $ .
But, from above calculation we can see that, $ 74g $ of $ Ca{(OH)_2} $ is equal to $ 1 $ mol of $ Ca{(OH)_2} $ , so, assuming sufficient amount of hydrochloric acid is present in the reaction, then $ 1 $ mol of $ CaC{l_2} $ will be,
 $ m = n \times Molar{\text{ }}mass = 1 \times (40 + 2 \times 35.5) $
Where, $ m \to $ mass of $ CaC{l_2} $ produced in the reaction
 $ n \to $ no of moles of $ CaC{l_2} $ produced in the solution
 $ m = 40 + 71 = 111g $
Hence, the total number of grams of $ CaC{l_2} $ that could be produced from $ 74g $ of $ Ca{(OH)_2} $ will be $ 111g $.

Note:
Before calculating the number of moles of $ Ca{(OH)_2} $ or $ CaC{l_2} $ , check whether the chemical equation is balanced or not. Since, a balanced chemical equation will give you a perfect mole-to-mole ratio. Also, try to remember the atomic weights of elements are correct as incorrect atomic weights can result in incorrect no. of moles being calculated.