Question

2. (7 pts.) Limiting Reactant and Theoretical Yield (7 pts.) 2Na (s) + Cl2 (g) - 2NaCl (s) You are given 84.9 g Na and 53.2 g Cl2. Find the limiting reactant, and calculate the theoretical yield in grams from the reaction shown above. (Box your answers!) Na = 22.990 g/mol; CI = 35.453 g/mol 22.990 mo lng & Appoy angl nach 229909x imoina Trimorol mol Nad mol 8Hana 2 molNG =0.265 35.453 g/mol x Imola x 2 mol Nach sangah 1 mol az wereana 1.22 35.453 4100% dox 58.28612 133 127%

Answer 1

1)

Molar mass of Na = 22.99 g/mol

mass(Na)= 84.9 g

use:

number of mol of Na,

n = mass of Na/molar mass of Na

=(84.9 g)/(22.99 g/mol)

= 3.693 mol

Molar mass of Cl2 = 70.9 g/mol

mass(Cl2)= 53.2 g

use:

number of mol of Cl2,

n = mass of Cl2/molar mass of Cl2

=(53.2 g)/(70.9 g/mol)

= 0.7504 mol

Balanced chemical equation is:

2 Na + Cl2 ---> 2 NaCl

2 mol of Na reacts with 1 mol of Cl2

for 3.693 mol of Na, 1.846 mol of Cl2 is required

But we have 0.7504 mol of Cl2

so, Cl2 is limiting reagent

Answer: Cl2

2)

Molar mass of NaCl,

MM = 1*MM(Na) + 1*MM(Cl)

= 1*22.99 + 1*35.45

= 58.44 g/mol

According to balanced equation

mol of NaCl formed = (2/1)* moles of Cl2

= (2/1)*0.7504

= 1.501 mol

use:

mass of NaCl = number of mol * molar mass

= 1.501*58.44

= 87.7 g

Answer: 87.7 g