Question

when 6 grams of zinc are dropped into excess hydrochloric acid, how many grams of zinc chloride will be produced?

Answer 1

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

Answer 2

Answer: The mass of zinc chloride produced will be 12.402 grams.

Step-by-step explanation:

To calculate the number of moles, we use the equation:

`\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}` .....(1)

• For zinc:

Given mass of zinc = 6 g

Molar mass of zinc = 65.38 g/mol

Putting values in above equation, we get:

`\text{Moles of zinc}=(6g)/(65.38g/mol)=0.091mol`

For the reaction of zinc and hydrochloric acid, the chemical equation follows:

`Zn(s)+2HCl(aq.)\rightarrow ZnCl_2(s)+H_2(g)`

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 =
`(1)/(1)* 0.091=0.091moles` 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:

`0.091=\frac{\text{Mass of zinc chloride}}{136.286g/mol}\\\\\text{Mass of zinc chloride}=12.402g`

Hence, the mass of zinc chloride produced will be 12.402 grams.