Question

Given 20.0 g Na and 29.5 g Cl2. Determine which is the limiting reactant. Wat theoretical yield in grams?

Answer 1

1)

Molar mass of Na = 22.99 g/mol

mass(Na)= 20.0 g

use:

number of mol of Na,

n = mass of Na/molar mass of Na

=(20 g)/(22.99 g/mol)

= 0.8699 mol

Molar mass of Cl2 = 70.9 g/mol

mass(Cl2)= 29.5 g

use:

number of mol of Cl2,

n = mass of Cl2/molar mass of Cl2

=(29.5 g)/(70.9 g/mol)

= 0.4161 mol

Balanced chemical equation is:

2 Na + Cl2 ---> 2 NaCl

2 mol of Na reacts with 1 mol of Cl2

for 0.8699 mol of Na, 0.435 mol of Cl2 is required

But we have 0.4161 mol of Cl2

so, Cl2 is limiting reagent

Answer: Cl2 is limiting reagent

2)

we will use Cl2 in further calculation

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)*0.4161

= 0.8322 mol

use:

mass of NaCl = number of mol * molar mass

= 0.8322*58.44

= 48.63 g

Answer: Theoretical yield = 48.6 g