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
-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
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.)
To solve this stoichiometry problem, we need to use mole definitions that relate to the given equation and the given units.
First, let's convert the 80.0 liters of oxygen gas to moles using the equation: 1 mole = 22.4 L. Since the given volume of oxygen gas is 80.0 L, we can calculate the number of moles of oxygen gas.
Next, we need to use the stoichiometric coefficients of the equation to find the mole ratio between oxygen gas (O2) and sodium chloride (NaCl). According to the given equation, there are 3 moles of oxygen gas produced for every 2 moles of sodium chloride. This mole ratio is necessary to calculate the moles of sodium chloride produced.
Finally, to find the grams of sodium chloride produced, we need to use the molar mass of sodium chloride (NaCl) which can be found using the molar mass of sodium (Na) and chlorine (Cl). According to the given mole definitions, the molar mass of NaCl is 58.5 g/mol.
So, for this stoichiometry problem, we would need the following mole definitions:
- 1 mole = 22.4 L = 22,400 mL = 22.4 dm3 = 22,400 cm3 of any gas at STP (to convert the given volume of oxygen gas to moles)
- 1 mole = 32.0 g O2 (to calculate the moles of oxygen gas)
- 1 mole = 58.5 g NaCl (to convert the moles of sodium chloride to grams)
Therefore, the mole definitions needed for this problem are:
- 1 mole = 22.4 L = 22,400 mL = 22.4 dm3 = 22,400 cm3 of any gas at STP
- 1 mole = 32.0 g O2
- 1 mole = 58.5 g NaCl