Answer:
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