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

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.