Respuesta :
Following reaction is initiated with Zn interacts with HCl;
Zn + 2HCl → ZnCl2 + H2
(1mol = 65.38 g) (1 mol = 136.28g)
Thus, 1 mol of Zn generates 1 mol of ZnCl2
Now, 6 g Zn ≡ 0.09177 mol.
Hence, 0.09177 mol of Zn generates 0.09117 mol of ZnCl2.
And 0.09117 mol of ZnCl2 ≡ 12.506 g
Thus, 6 g of Zn when dropped into excess HCl, it will produce 12.506 g of ZnCl2
Zn + 2HCl → ZnCl2 + H2
(1mol = 65.38 g) (1 mol = 136.28g)
Thus, 1 mol of Zn generates 1 mol of ZnCl2
Now, 6 g Zn ≡ 0.09177 mol.
Hence, 0.09177 mol of Zn generates 0.09117 mol of ZnCl2.
And 0.09117 mol of ZnCl2 ≡ 12.506 g
Thus, 6 g of Zn when dropped into excess HCl, it will produce 12.506 g of ZnCl2
Answer: The mass of zinc chloride produced will be 12.402 grams.
Explanation:
To calculate the number of moles, we use the equation:
[tex]\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}[/tex] .....(1)
- For zinc:
Given mass of zinc = 6 g
Molar mass of zinc = 65.38 g/mol
Putting values in above equation, we get:
[tex]\text{Moles of zinc}=\frac{6g}{65.38g/mol}=0.091mol[/tex]
For the reaction of zinc and hydrochloric acid, the chemical equation follows:
[tex]Zn(s)+2HCl(aq.)\rightarrow ZnCl_2(s)+H_2(g)[/tex]
Hydrochloric acid is given in excess, so it is considered as an excess reagent. Zinc is considered as a limiting reagent because it limits the formation of product.
By Stoichiometry of the reaction:
1 mole of zinc metal produces 1 mole of zinc chloride.
So, 0.091 moles of zinc metal will produce = [tex]\frac{1}{1}\times 0.091=0.091moles[/tex] of zinc chloride.
Now, calculating the mass of zinc chloride produced, we use equation 1.
Moles of zinc chloride = 0.091 moles
Molar mass of zinc chloride = 136.286 g/mol
Putting values in equation 1, we get:
[tex]0.091=\frac{\text{Mass of zinc chloride}}{136.286g/mol}\\\\\text{Mass of zinc chloride}=12.402g[/tex]
Hence, the mass of zinc chloride produced will be 12.402 grams.
