Question

How many grams of zinc metal will react completely with 5.4 liters of 2.8 M HCl? Show all of the work needed to solve this problem. Zn (s) + 2 HCl (aq) yields ZnCl2 (aq) + H2 (g)

Answer 1

Answer : The amount of zinc metal react will be, 494.27 grams.

Solution : Given,

Volume of HCl solution = 5.4 L

Molarity of HCl solution = 2.8 M

Molar mass of zinc (Zn) = 65.38 g/mole

First we have to calculate the moles of HCl.

`\text{Molarity of }HCl=\frac{\text{Moles of }HCl}{\text{Volume of solution}}`

`2.8M=\frac{\text{Moles of }HCl}{5.4L}`

`\text{Moles of }HCl=15.12moles`

Now we have to calculate the moles of Zn.

The given balanced chemical reaction is,

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

From the balanced reaction, we conclude that

As, 2 moles of
`HCl` react to with 1 mole of Zn

So, 15.12 moles of
`HCl` react to with
`(15.12)/(2)=7.56` moles of Zn

Now we have to calculate the mass of Zn.

`\text{Mass of }Zn=\text{Moles of }Zn* \text{Molar mass of }Zn`

`\text{Mass of }Zn=(7.56mole)* (65.38g/mole)=494.27g`

Therefore, the amount of zinc metal react will be, 494.27 grams.

Answer 2

The reaction is
Zn (s) + 2 HCl (aq) ----> ZnCl2 (aq) + H2 (g)
which is already balanced
5.4 L of 2.8 M HCl contains
5.4 L (2.8 M) = 15.12 moles HCl
The amount of Zinc that will react completely with the acid is
15.12 mol HCl (1 mol Zn / 2 mol HCl) (65 g Zn/1 mol Zn) = 491.4 g Zn