Question

How many grams of maximum zinc chloride will be formed from a reaction between $ 6.5{\text{ }}g $ of zinc $ 3.2{\text{ }}g $ of $ HCl $ ?
(atomic weight of $ Zn = 65 $ and $ Cl = 35.5 $ )
 $
  A. 5.96{\text{ }}g \\
  B. 3.97{\text{ }}g \\
  C. 7.97{\text{ }}g \\
  D. 2.97{\text{ }}g \\
  $

Answer 1

Hint: To calculate number of moles we have a formula which is $ number{\text{ }}of{\text{ }}moles = \dfrac{{given{\text{ }}mass}}{{molar{\text{ }}mass}} $
To find the answer to this knowing the complete balancing reaction is important. Limiting reactant- a reactant that is consumed when the chemical reaction is completed.

Complete answer:
 We are given the weight of zinc $ = 6.5{\text{ }}g $ and of HCl $ = 3.2{\text{ }}g $ . We need to find the amount of zinc chloride formed. First let us look how the reaction will occur:
 $ Zn + 2HCl \to ZnC{l_2} + {H_2} $
Now let us try to solve it step by step. First we will calculate a mole of a $ Zn $ and $ HCl $ to know which one will be the limiting reagent. For that we will calculate the moles of both $ Zn $ and $ HCl $
For zinc:
 $ number{\text{ }}of{\text{ }}moles = \dfrac{{given{\text{ }}mass}}{{molar{\text{ }}mass}} $
Given mass $ = 6.5g $ Molar mass $ = 65.3\dfrac{g}{{mol}} $
 $ number{\text{ }}of{\text{ }}moles = \dfrac{{6.5}}{{65.3}} = 0.99mol $
Now for $ HCl $ :
Given mass $ = 3.2g $ Molar mass $ = 36.5\dfrac{g}{{mol}} $
 $ number{\text{ }}of{\text{ }}moles = \dfrac{{3.2}}{{36.5}} = 0.087mol $
Now in the reaction we have seen that $ 2 $ moles of $ HCl $ reacted with $ 1 $ mole of $ Zn $ , so it means that
 $ 0.087mol $ Of $ HCl $ will react with = $ \dfrac{1}{2} \times 0.087 = 0.0435moles $ of $ Zn $ .
We can see here the number of moles of $ HCl $ is less than the number of moles of $ Zn $ , thus $ HCl $ will be our limiting reagent.
Next we will calculate the mass of zinc chloride; for calculating this we will use the same above formula, but we will interchange it slightly. So by interchanging, the formula comes out be as:
  $ mass = number{\text{ }}of{\text{ }}moles \times molar{\text{ }}mass $
Now for the moles of $ ZnC{l_2} $ , we know that from the reaction $ 2 $ moles of $ HCl $ produces $ 1 $ mole of $ ZnC{l_2} $ so, $ 0.087mol $ Of $ HCl $ will produce $ = \dfrac{1}{2} \times 0.087 = 0.0435moles $ of $ ZnC{l_2} $
Molar mass of $ ZnC{l_2} $ $ = 136.286{\text{ }}\dfrac{g}{{mol}} $
Thus mass of $ ZnC{l_2} $ is : $ mass = number{\text{ }}of{\text{ }}moles \times molar{\text{ }}mass $
 $ mass = 0.0435 \times 136.286 $
 $ = 5.9284g $
Hence, the mass of $ ZnC{l_2} $ formed will be $ 5.9284g $
Therefore the correct option is A. $ 5.96g $

Note:
Atomic weight and molar mass are similar. The only difference between them is that atomic weight is the average mass of an element and is expressed in atomic mass unit and molar mass is the mass of a given substance(g) divided by the amount of substance(mol).