Question

1. What is the mass of 22.4 L of H2 at STP?

Answer 1

1.
n = m/2.016 g/mol = V/22.4 l
V = 22.4 l
Hence,
m = 2.016 g ≈ 2.02 g
B.) 2.02 g

2.
Balancing the equation, we get,
Mg + 2HCl → MgCl2 + H2
PV = nRT
V = nRT/P
V = 1.5 mol × 8.314 atm/(mol.K) × 298 K/(0.96 atm)
V = 3871.206
V ≈ 38.2 l

Answer 2

Explanation:

1.    According to Avogadro's law 1 mole of every gas occupies 22.4 L. Therefore, it is given that volume is 22.4 L so, this means that number of moles equal to one.

Molar mass of hydrogen gas $(H_{2})$ is 2.02 g/mol.

Hence, calculate the mass as follows.

    No. of moles = $\frac{mass}{molar mass}$

or,        mass = $No. of moles \times molar mass$

                     = $1 \times 2.02 g/mol$

                     = 2 g

Therefore, mass of 22.4 L of $H_{2}$ at STP is 2.02 grams.

2.   The given data is as follows.

         P = 0.960 atm,                T = 298 K

         n = 3.00 moles,               V = ?

         R = 0.0821 atm L/K mol

So, calculate the volume using ideal gas equation as follows.

                   PV = nRT

                     V = $\frac{nRT}{P}$

                         = $\frac{3 mol \times 0.0821 atm L/K mol \times 298 K}{0.960 atm}$

                         = 76.455 L

Thus, we can conclude that volume of the hydrogen gas produced is 76.455 L.