Answer:
Limiting Reagent: NaHCO₃
Amount of NaCl: 69.6 g NaCl
Mass of Excess Reagent: 56.6 g HCl
Step-by-step explanation:
To find the mass of NaCl, you need to (1) convert mass reactant to moles reactant (via molar mass), then (2) convert moles reactant to moles NaCl (via mole-to-mole ratio from equation coefficients), and then (3) convert moles NaCl to mass NaCl (via molar mass).
Molar Mass (NaHCO₃):
22.990 g/mol + 1.008 g/mol + 12.011 g/mol + 3(15.998 g/mol)
Molar Mass (NaHCO₃): 84.003 g/mol
Molar Mass (HCl): 1.008 g/mol + 35.453 g/mol
Molar Mass (HCl): 36.461 g/mol
Molar Mass (NaCl): 22.990 g/mol + 35.453 g/mol
Molar Mass (NaCl): 58.443 g/mol
100 g NaHCO₃ 1 mole 1 mole NaCl 58.443 g
------------------------ x ------------------ x --------------------------- x ---------------- =
84.003 g 1 mole NaHCO₃ 1 mole
= 69.6 g NaCl
100 g HCl 1 mole 1 mole NaCl 58.443 g
---------------- x ------------------ x ----------------------- x ---------------- =
36.461 g 1 mole HCl 1 mole
= 160 g NaCl
Since NaHCO₃ results in the smaller amount of product, it is the limiting reagent. In other words, it is used up before the HCl has the chance to completely react. Therefore, the actual amount of NaCl produced is 69.6 grams.
To find the mass of the excess reagent, you need to calculate the amount of HCl actually used in the reaction. Then, you need to subtract that value from the total amount of HCl.
69.6 g NaCl 1 mole 1 mole HCl 36.461 g
--------------------- x ------------------ x ----------------------- x ---------------- =
58.443 g 1 mole NaCl 1 mole
= 43.4 g HCl
Amount Given - Amount Reacted = Mass Excess
100 g HCl - 43.4 g HCl = 56.6 g HCl