Exercise 6 | Chemical Calculations

How many grams of NaCl (s) can be produced from 5.00 grams of Na (s) with an excess of Cl2 (g)?

What is the minimum mass of Cl2 (g) required?

Equation: 2 Na (s) + Cl2 (g) → 2 NaCl (s)

2 moles of Na form 2 moles of NaCl ⇒ n consumed Na = n produced NaCl


n consumed Na = mNaMNa ⇒ n consumed Na = 5.0023.0 = 2.17 x 10-1 mol

⇒ n produced NaCl = 2.17 x 10-1 mol

n produced NaCl = nNaCl = mNaClMNaCl ⇒ mNaCl = n produced NaCl x MNaCl ⇒ mNaCl = 12.7 g


1 mole of Cl2 is necessary to form 2 moles of NaCl

⇒ nrequired Cl2 = nproduced NaCl2

nrequired Cl2 = 1.09 x 10-1 mol

nrequired Cl2 = mCl2MCl2 ⇒ mCl2 = n required Cl2 x MCl2 ⇒ mCl2 = 7.71 g