Question

Consider the reaction between the zinc metal (Zn) and hydrochloric acid (HCl),

2HCl (aq) + Zn (s) → ZnCl₂ (aq) + H₂ (g).

How many grams of hydrogen gas (H₂) are possible when 0.500 grams of Zn is added to 250 mL of a 0.123 M HCl solution? (Answer: 0.0154 grams of H₂)

Please give a step by step explanation. Answer is given above

Answer 1

volume of HCl, V = 250 mL

= 0.25 L

we have below equation to be used:

number of mol in HCl,

n = Molarity * Volume

= 0.123*0.25

= 3.075*10^-2 mol

Molar mass of Zn = 65.38 g/mol

mass of Zn = 0.5 g

we have below equation to be used:

number of mol of Zn,

n = mass of Zn/molar mass of Zn

=(0.5 g)/(65.38 g/mol)

= 7.648*10^-3 mol

we have the Balanced chemical equation as:

2 HCl + Zn ---> H2 + ZnCl2

2 mol of HCl reacts with 1 mol of Zn

for 3.075*10^-2 mol of HCl, 1.537*10^-2 mol of Zn is required

But we have 7.648*10^-3 mol of Zn

so, Zn is limiting reagent

we will use Zn in further calculation

Molar mass of H2 = 2.016 g/mol

From balanced chemical reaction, we see that

when 1 mol of Zn reacts, 1 mol of H2 is formed

mol of H2 formed = (1/1)* moles of Zn

= (1/1)*7.648*10^-3

= 7.648*10^-3 mol

we have below equation to be used:

mass of H2 = number of mol * molar mass

= 7.648*10^-3*2.016

= 0.0154 g

Answer: 0.0154 g