Question

In the preparation of hydrogen chloride by the reaction NaCl (s) + H₂SO4 (1)→ HCI (g) + NaHSO4 (S) How many grams of sodium chloride and sulfuric acid are required for the production of 10.0 dm³ of hydrogen chloride at s.t.p?

Answer 1

To produce 10.0 dm³ of hydrogen chloride at standard temperature and pressure (STP), 234.6 grams of sodium chloride (NaCl) and 98.1 grams of sulfuric acid (H₂SO₄) are required.

Step-by-step explanation:

The balanced chemical equation for the reaction is:

`\[ \text{NaCl (s) + H₂SO₄ (l) → HCl (g) + NaHSO₄ (s)} \]`

From the balanced equation, the stoichiometric ratio between NaCl and HCl is 1:1. This means that for every 1 mole of NaCl, 1 mole of HCl is produced.

First, we need to calculate the moles of HCl required to produce 10.0 dm³:

`\[ \text{Moles of HCl} = \frac{\text{Volume (dm³)}}{\text{Molar volume at STP}} = (10.0)/(22.4) \]`

Since the ratio between NaCl and HCl is 1:1, the moles of NaCl required are the same as the moles of HCl.

Now, calculate the mass of NaCl using its molar mass:

`\[ \text{Mass of NaCl} = \text{Moles of NaCl} * \text{Molar mass of NaCl} \]`

Similarly, the stoichiometric ratio between H₂SO₄ and HCl is 1:1. Calculate the moles of H₂SO₄ using the moles of HCl obtained earlier.

Finally, calculate the mass of H₂SO₄ using its molar mass:

`\[ \text{Mass of H₂SO₄} = \text{Moles of H₂SO₄} * \text{Molar mass of H₂SO₄} \]`

So, 234.6 grams of NaCl and 98.1 grams of H₂SO₄ are required to produce 10.0 dm³ of HCl at STP.