Question

For the following stoichiometry problem, select ALL of the mole definitions you would need to solve the problem.

Problem: Given the following equation:

2NaClO3 --> 2NaCl + 3O2

How many grams of sodium chloride, NaCl, are produced when 80.0 liters of oxygen gas are produced?
Select all that apply:
-1 mole = 22.4 L = 22,400 mL = 22.4 dm3 = 22,400 cm3 of any gas at STP

-1 mole = 6.02 x 1023 particles

-1 mole = 32.0 g O2

-1 mole = 58.5 g NaCl

-1 mole = 106.5 g NaClO3

Answer 1

-1 mole = 58.5 g NaCl
-1 mole = 22.4 L = 22,400 mL = 22.4 dm3 = 22,400 cm3 of any gas at STP

Answer 2

Converting given data to moles typically uses ...
grams to moles => moles = mass(g) given/formula wt(g/mole)
particles to moles => moles = particle numbers given/Avogadro's Number
volume at STP (Liters) to moles => moles = Volume(L) given/22.4l/mole
solutions data (Molarity(M) & Volume(L) given) => moles = Molarity(M) x Volume(L)
energy values to moles => moles = energy value given/molar heat of rxn

For the question/problem given...
2NaClO₃ => 2NaCl + 3O₂
Assuming reaction given is at STP, then moles O₂ produced = 80.0L/22.4L/mole = 3.57 moles O₂
From equation stoichiometry, moles NaCl produced along with the 3.57 moles O₂ = 2/3(3.57)moles NaCl = 2.38 moles NaCl = 2.38 mole(58g/mole)NaCl = 138.1g NaCl ~ 138g (3 sig. figs. Based on the 80.0L Oxy given.)