Question

Find the mass of sodium required to reduce 6.58 L of hydrogen gas at 32°C and 895 mm, when sodium reacts with hydrochloric acid. (Hint: you first need to write a balanced chemical equation.)

Answer 1

Mass = 14.72 g

Step-by-step explanation:

Given data:

Volume of hydrogen = 6.58 L

Temperature of gas = 32°C (32+273 = 305 k)

Pressure of gas = 895 mmHg (895/760 = 1.2 atm)

Mass of sodium required = ?

Solution:

Chemical equation:

2HCl + 2Na → 2NaCl + H₂

Number of moles of hydrogen:

PV = nRT

1.2 atm × 6.58 L = n× 0.0821 atm.L/mol.K×305 K

7.9 atm.L = n× 25.0 atm.L/mol

n = 7.9 atm.L / 25.0 atm.L/mol

n = 0.32 mol

Now we will compare the moles of hydrogen with sodium.

H₂ : Na

1 : 2

0.32 : 2/1×0.32 = 0.64 mol

Mass of sodium:

Mass = number of moles × molar mass

Mass = 0.64 mol × 23 g/mol

Mass = 14.72 g