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)

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

Answer Link

BarrettArcher BarrettArcher · Answer 2 · 2019-02-19T19:45:57+01:00

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.

[tex]\text{Molarity of }HCl=\frac{\text{Moles of }HCl}{\text{Volume of solution}}[/tex]

[tex]2.8M=\frac{\text{Moles of }HCl}{5.4L}[/tex]

[tex]\text{Moles of }HCl=15.12moles[/tex]

Now we have to calculate the moles of Zn.

The given balanced chemical reaction is,

[tex]Zn(s)+2HCl(aq)\rightarrow ZnCl_2(aq)+H_2(g)[/tex]

From the balanced reaction, we conclude that

As, 2 moles of [tex]HCl[/tex] react to with 1 mole of Zn

So, 15.12 moles of [tex]HCl[/tex] react to with [tex]\frac{15.12}{2}=7.56[/tex] moles of Zn

Now we have to calculate the mass of Zn.

[tex]\text{Mass of }Zn=\text{Moles of }Zn\times \text{Molar mass of }Zn[/tex]

[tex]\text{Mass of }Zn=(7.56mole)\times (65.38g/mole)=494.27g[/tex]

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