Question

How would I do this problem: Sodium chlorate (NaClO3) can be used as the active ingredient in some weedkillers. When it decomposes, it forms sodium chloride and oxygen gas. Calculate the mass of sodium chlorate needed to produce 25.0g of oxygen gas?

Answer 1

mass of sodium chlorate required = 55.4 g (3 sig figs)

Step-by-step explanation:

In chemical equations, unless stated otherwise, the reactants and products will theoretically always remain in stoichiometric ratios.

The stoichiometry of a reaction is the relationship between the relative quantities of products and reactants, typically a ratio of whole integers.

Consider the following chemical reaction: aA + bB ⇒ cC + dD.

The stoichiometry of reactants to products in this reaction is the ratio of the coefficients of each species: a : b : c : d.

Now let's apply this knowledge to the question to be attempted:

first, we can start by writing out a balanced chemical equation, with states.

2NaClO₃(s) ⇒ 2NaCl(s) + 3O₂(g)

Hence, the stoichiometry of this reaction is 2 : 2 : 3. This means that the moles of NaClO₃ = moles of NaCl = 2/3 × moles of O₂.

Now if we can convert the mass of O₂ into moles, then we can calculate the moles of NaClO₃, and thus the mass of NaClO₃.

Now, to speed things up a bit, we will use the formula n = m/M, where

• number of moles: symbol (n), unit (mol)
• molar mass: symbol (M), unit (g/mol)
• mass present: symbol (m), unit (g)

Therefore, n(O₂) = m/M = 25.0/(16.00×2) = 0.78125 mol

n(NaClO₃) = 2/3 × n(O₂) = 2/3 × 0.78125 = 0.52083 mol

Finally, rearranging the original formula to find mass,

m(NaClO₃) = n×M = 0.52083×(22.99+35.45+16.00×3) = 55.4 grams