Question

How many grams of zinc metal will react completely with 8.2 liters of 3.5 M HCl? Show all of the work needed to solve this problem. Zn (s) + 2HCl (aq)yields ZnCl2 (aq) + H2 (g)

Answer 1

Answer: 938.203 grams of zinc metal will react completely with 8.2 liters of 3.5 M HCl.

Explanation:

Molarity of the HCl solution = 3.5 M = 3.5 mol/L

Volume of the HCl solution = 8.2 l

$Molarity=3.5 mol/L=\frac{\text{Moles of HCl}}{\text{Volume in liters}}$

Moles of HCl = $3.5 mol/L\times 8.2 L=28.7 moles$

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

According to reaction 2 moles of the HCl reacts with 1 mole of Zn metal.

Then 28.7 moles of HCl will react with:

$\frac{1}{2}\times 28.7$ moles of Zinc metal that is 14.35 moles.

Mass of the zinc metal required to completely react with given HCl solution is:

Mass of Zinc = Moles of zinc × Atomic Mass of zinc

= 14.35 mol × 65.38 g/mol = 938.203 g

938.203 grams of zinc metal will react completely with 8.2 liters of 3.5 M HCl.

Answer 2

Answer: 1951.6 g

Explanation:

Molarity : It is defined as the number of moles of solute present in one liter of solution.

Moles of HCl=$Molarity\times {\text { Volume in L}$

Moles of HCl=$3.5M\times 8.2L=28.7moles$

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

From the given balanced chemical equation,

2 moles of hydrogen chloride react with 1 mole of Zinc.

28.7 moles hydrogen chloride react with=$\frac{1}{2}\times 28.7=14.35$ moles of Zinc.

Mass of $ZnCl_2=Moles\times {\text{Molar mass}}$  

Mass of $ZnCl_2=14.35\times 136g/mol=1951.6g$  

Thus 1951.6 g of zinc metal will react completely with 8.2 liters of 3.5 M HCl.