Question

Определите объем хлороводорода при нормальных условиях, который можно получить действием концентрированной серной кислоты на хлорид натрия массой 11 г.​

Answer 1

N(NaCl)=11,7:58,5= 0,2 моль Mr(NaCl) =58,5

NaCl + H2SO4 = NaHSO4 + HCL
1 моль----------------------------1 моль
0,2 моль-----------------------0,2 моль
V(HCl) =0,2*22,4= 4,48 (л) Vm=22,4 л/моль при н. у.

Answer 2

Calculating the Volume of Hydrochloric Acid Produced

To calculate the volume of hydrochloric acid produced when concentrated sulfuric acid reacts with 11 grams of sodium chloride, we can use the following information from the search results:

The balanced chemical equation for the reaction is: 2HCl + Na2SO4 → 2NaCl + H2SO4

From the search results, we can find the molar mass of sodium chloride: The molar mass of NaCl is 58.44 g/mol.

The molar ratio between sodium chloride and hydrochloric acid is 1:2: 1 mole of NaCl produces 2 moles of HCl.

Using this information, we can calculate the amount of hydrochloric acid produced and then convert it to volume.

Calculation Steps

1. Calculate the moles of sodium chloride: - Moles = Mass / Molar mass - Moles = 11 g / 58.44 g/mol - Moles ≈ 0.188 moles.

2. Calculate the moles of hydrochloric acid produced: - Moles of HCl = 2 * Moles of NaCl - Moles of HCl = 2 * 0.188 moles - Moles of HCl ≈ 0.376 moles.

3. Convert moles of hydrochloric acid to volume at standard conditions (STP): - 1 mole of any gas at STP occupies 22.4 liters. - Volume of HCl = Moles of HCl * 22.4 L/mol - Volume of HCl ≈ 8.4224 liters.

Therefore, the volume of hydrochloric acid produced when concentrated sulfuric acid reacts with 11 grams of sodium chloride at standard conditions is approximately 8.4224 liters.