Chem HWK due TODAY Help please ?
Consider the reaction 2 Na+ Cl2 -> 2 NaCl(s) which contains 53.2 g of Na and 65.8 g of Cl2.
a. perform the three conversions required to determine the mass of NaCl that would be made if there were more than enough Cl2
b. Beginning with the mass of Cl2 available, perform the three conversions required to determine the mass of NaCl that would be made if there were more than enough Na.
c. Which calculation predicts the least amount of product?
d. What is the limiting reactant? What is the theoretical yield?
- micatkieLv 72 months agoFavorite Answer
Molar mass of Na = 23.0 g/mol
Molar mass of Cl₂ = 35.5× 2 = 71.0 g/mol
Molar mass of NaCl = 23.0 + 35.5 = 58.5 g/mol
2Na + Cl₂ → 2NaCl
Mole ratio Na : Cl₂ : NaCl = 2 : 1 : 2
(53.2 g Na) × (1 mol Na / 23.0 g Na) × (2 mol NaCl / 2 mol Na) × (58.5 g NaCl / 1 mol NaCl)
= 135 g NaCl
(65.8 g Cl₂) × (1 mol Cl₂ / 71.0 g Cl₂) × (2 mol NaCl / 1 mol Cl₂) × (58.5 g NaCl / 1 mol NaCl)
= 108 g NaCl
The calculation in b, beginning with the mass of Cl₂ available, gives the least amount of product.
The limiting reactant is Cl₂.
The theoretical yield is 108 g.