Question

Hydrochloric acid can be prepared by the following reaction: 2NaCl(s) + H₂SO₄(aq) → 2HCl(g) + Na₂SO₄(s) what mass of hcl can be prepared from 2.00 mol h2so4 and 150. g nacl?

Answer 1

2.00 moles of H2SO4 will react to produce 145.84g of HCl. To determine the mass of HCl prepared from H2SO4 and NaCl, follow the stoichiometric relationship. H2SO4 is the limiting reagent, we need to find out the limiting reagent by comparing the mole ratio based on provided quantities.

Step-by-step explanation:

The mass of HCl that can be prepared from 2.00mol of H2SO4 and 150g of NaCl can be determined by stoichiometry. The reaction

2NaCl(s) + H2SO4(aq) → 2HCl(g) + Na2SO4(s)

shows that 2 moles of NaCl react with 1 mole of H2SO4 to produce 2 moles of HCl. First, we need to find out the limiting reagent by comparing the mole ratio based on provided quantities. For NaCl: Mass of NaCl = 150g Molar mass of NaCl = 58.44g/mol Moles of NaCl = Mass of NaCl / Molar mass of NaCl = 150g / 58.44g/mol = 2.57 moles.

Since the reaction ratio is 2:1, we need twice the moles of NaCl to react with H2SO4, which means only 2.00 moles of H2SO4 can react with 2.00 moles of NaCl. Therefore, H2SO4 is the limiting reagent.

We can use its moles to calculate the amount of HCl produced:

Moles of HCl produced = 2 × Moles of H2SO4 = 2 × 2.00moles = 4.00moles

Molar mass of HCl = 36.46g/mol

Mass of HCl produced = Moles of HCl × Molar mass of HCl

= 4.00moles × 36.46g/mol = 145.84g HCl.