1.2 x 10² L of oxygen is formed if 77.37 L of water is produced.
The density of water at STP = 0.99 g/ ml but we will assume 1 g/ml
Determine the mass of water ([tex]H_{2}[/tex]o) = 77.37 L × [tex]\frac{1000 ml}{1l}[/tex] × [tex]\frac{1g}{ml}[/tex]
The mass of water = 77370 g
Convert mass into moles = 77370 [tex]H_{2}[/tex]o × [tex]\frac{1mol}{18.01528} H _{2}O[/tex]
= 4294.6876 mol of [tex]H_{2} O[/tex]
Use stoichiometry to find how many mols of oxygen [tex]o_{2}[/tex] were used:
= 4294.6876 mol of [tex]H_{2} O[/tex] × [tex]\frac{13 mol }{10 mol}[/tex]
= 5583.0939 mol [tex]o_{2}[/tex]
Convert mols to mass = 5583.0939 mol ×[tex]\frac{32 g}{mol }[/tex]
= 178659.0059 g [tex]o_{2}[/tex]
Convert mass into liters
The density of [tex]o_{2}[/tex] at STP = [tex]\frac{1.439}{ml}[/tex]
Density of [tex]o_{2}[/tex] = 1.2 × [tex]10^{2}[/tex]
Hence,1.2 x 10² L of oxygen is formed if 77.37 L of water is produced.
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