Answer to Question #138468 in General Chemistry for Javier

Question

Mg metal reacts with HCl to produce hydrogen gas:
Mg(s) + 2HCl(aq) → MgCl2 (aq) + H2(g)


What volume of hydrogen at STP is released when 8.35 g Mg reacts? (Assume that the amount of HCl is enough for the reaction.)

Answer 1

Mg metal reacts with HCl to produce hydrogen gas:

Mg(s) + 2HCl(aq) → MgCl₂ (aq) + H₂(g)

What volume of hydrogen gas at STP is released when 8.35g Mg reacts? (Assume that the amount of HCl is enough for the reaction.)

Answer:

The given reaction is

Mg(s) + 2HCl(aq) → MgCl₂ (aq) + H₂ (g)

The reaction is already balanced for each element.

Mass of Mg reacted = 8.35g.

Step 1 : Convert 8.35g of Mg to moles.

We will convert 8.35g of Mg to 'mol of Mg' using atomic mass of Mg.

Atomic mass of Mg = 24.305g/mol

moles of Mg = 8.35g of Mg * 1mol Mg/24.305g Mg = 0.3436 mol Mg

Step 2 : To find 'mol of H₂' formed from 0.3436 mol of Mg

The reaction is

1 Mg(s) + 2HCl(aq) → 1 MgCl₂ (aq) + 1H₂(g)

The mole ratio of Mg and H₂ in this reaction is 1:1.

So the moles of H₂ formed = 0.3436 mol Mg * 1mol H₂ / 1mol Mg = 0.3436mol H₂ .

Step 3 : To find the volume of hydrogen released at STP.

At STP, P = 1atm and T = 273.15K.

We know, mole of H₂ = 0.3436mol and

Gas constant, R = 0.08206L-atm/mol-K.

Plug this information in the ideal gas equation, PV=nRT and solve it for V.

1atm * V = 0.3436mol * 0.08206L-atm/mol-K * 273.15K

1atm * V = 7.7017 L-atm

divide both the side by 1atm, we have

V = 7.7017L-atm / 1atm = 7.702L.

The quantity given in the question are in 3 significant figure, so we must round up our

final answer to 3 significant figure.

Hence Volume of H₂ released = 7.70L