Chemistry question (limiting reagent).?

Molar mass of NaCl = (22.99 + 35.45) g/mol = 58.44 g/mol
Molar mass of H₂SO₄ = 1.01×2 + 32.06 + 16.00×4 = 98.08 g/mol

NaCl + H₂SO₄ → HCl + Na₂SO₄

Initial number of moles of NaCl = (5.00 g) / (58.44 g/mol) = 0.0856 mol
Initial number of moles of H₂SO₄ = (10.00 g) / (98.08 g/mol) = 0.102 mol

According to the equation, mole ratio NaCl : H₂SO₄ = 1 : 1
As (Initial number of moles of NaCl) < (Initial number of moles of H₂SO₄),
NaCl is the limiting reagent.

"Sodium bisulfate" is NaHSO4

NaCl + H2SO4 -> HCI + Na2SO4 <<< skeleton equation

2NaCl + H2SO4 -> 2HCI + Na2SO4 <<< balanced equation
... select any product to use .. I will use HCl
Molar mass NaCl = 58.5
5g X (1mole/58.5g) X (2mole HCl / 2moleNaCl) = 0.085 mole HCl

molar mass H2SO4 = 98.1
10g X (1mole/98.1g) X (2mole HCl / 1 mole H2SO4) = 0.204 mole HCl

smallest answer is corrwxr for the product .. since NaCl was the reactant that gave the smallest answer, then it is the limiting reactant

NaCl + H2SO4 -> HCI +Na2SO4

The coefficients in a balanced equation determine the ratio of reactants to moles of products. Since all the coefficients are one, the number of moles of salt must be equal to the number of moles of sulfuric acid. To determine the number of moles, divide the mass by the mass of one mole.

NaCl = 23 + 35.5 = 58.5 g
n = 5 ÷ 58.5
This is approximately 0.085 moles.

H2SO4 = 2 + 32.1 + 4 * 16 = 98.1 g
n = 10 ÷ 98.1
This is approximately 0.102 moles.

In this problem, the limiting reagent is one with the least number of moles. This is salt.