Step 1
The reaction must be completed and balanced:
4NH3 (g) + 9O2 (g) => 4NO3 + 6H2O (g)
Reactants: NH3 and O2
Products: NO3 and H2O
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Step 2
Information provided:
STP conditions, therefore:
1 mole of gas (NH3 in this case) = 22.4 L (volume)
5.3 g of O2
Liters => L
Grams => g
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Information needed:
The molar mass of O2 = 32.0 g/mol
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Step 3
By stoichiometry,
1 mole NH3 = 22.4 L NH3
1 mole O2 = 32.0 g O2
Procedure: We will work with volume and mass
4NH3 (g) + 9O2 (g) => 4NO3 + 6H2O (g)
4 x 22.4 L NH3 ---------- 9 x 32.0 g O2
X ----------- 5.3 g O2
X = 5.3 g O2 x 4 x 22.4 L NH3/9 x 32.0 g O2 = 1.65 L approx.
Answer: 1.65 L of NH3