1. How many moles of oxygen gas are needed to form 21.8 liters of water vapor?
1 answer:
0.781 moles
Explanation:
We begin by balancing the chemical equation;
O₂ (g) + 2H₂ (g) → 2H₂O (g)
21.8 Liters = 21.8 Kgs
To find how many moles are in 28.1 Kg H₂O;
Molar mass of H₂O = 18 g/mol
28.1/18
= 1.56 moles
The mole ratio between water vapor and oxygen is;
1 : 2
x : 1.56
2x = 1.56
x = 1.56 / 2
x = 0.781
0.781 moles
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