When 22.0 g NaCl and 21.0 g H₂SO₄ are mixed and react according to the equation below, which is the limiting reagent
2NaCl + H₂SO₄ $\rarr$ Na₂SO₄ + 2HCl

Question

When 22.0 g NaCl and 21.0 g H₂SO₄ are mixed and react according to the equation below, which is the limiting reagent
2NaCl + H₂SO₄

\rarr

Na₂SO₄ + 2HCl

Quizplus · Accepted Answer

The chemical formulae seem to be incorrectly formatted, but based on the stoichiometry provided in the equation (2NaCl + 1HS... -> NaS... + 2HCl), and given masses, we can determine the limiting reagent by calculating the moles of each reactant and comparing them to the stoichiometric ratios. NaCl has a molar mass of approximately 58.44 g/mol, so 22.0 g corresponds to about 0.376 moles. Without the correct formula for the other reactant, we can't calculate its moles precisely, but since the equation requires twice as much NaCl as the other reactant, and given that both reactants are provided in similar masses, NaCl is likely the limiting reagent because it is needed in a greater amount according to the stoichiometry of the reaction.

When 220 G NaCl and 21 →\rarr→ Na2SO4 + 2HCl A) NaCl B) H2SO4 C) Na2SO4 D)

When 220 G NaCl and 21 $\rarr$ Na₂SO₄ + 2HCl
A) NaCl
B) H₂SO₄
C) Na₂SO₄
D)