Step 1
The equation provided:
4 Na (s) + O2 (g) → 2 Na2O (s) (completed and balanced)
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Step 2
Information provided:
28.5 g of Na
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Information needed:
The molar masses of:
Na2O = 61.9 g/mol
Na = 23.0 g/mol
(The periodic table is useful here)
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Step 3
Procedure:
By stoichiometry,
4 Na (s) + O2 (g) → 2 Na2O (s)
4 x 23.0 g Na -------------- 2 x 61.9 g Na2O (given by the reaction)
28.5 g Na -------------- X
X = (28.5 g Na x 2 x 61.9 g Na2O)/4 x 23.0 g Na
X = 38.4 g
Answer: 38.4 g of Na2O produced
