Answer:
256 g
Step-by-step explanation:
a) Balanced equation
We know we will need an equation with masses and molar masses, so let’s gather all the information in one place.
M_r: 2.016 32.00 18.02
2H₂ + O₂ ⟶ 2H₂O
m/g: 32.3 288
(b) Moles of H₂O
n = 288 g H₂O × (1 mol H₂O/18.02 g H₂O)
= 15.98 mol H₂O
(c) Moles of O₂
The molar ratio is (2 mol H₂O /1 mol O₂)
n = 15.98 mol H₂O × (1 mol O₂ /2 mol H₂O)
= 7.991 mol O₂
(d) Mass of O₂
m = 7.991 mol O₂× (32.00 g O₂/1 mol O₂)
= 256 g O₂
The reaction requires 256 g O₂.
<em>Note</em>: If you base your answer on the mass of <em>hydrogen</em>, you get the same result.
Mass of O₂ = 32.3 g H₂ × 1 mol H₂/2.016 g H₂
= 16.02 mol H₂ × 1 mol O₂ /2 mol H₂
= 8.011 mol O₂ × 32.00 g O₂/1 mol O₂
= 256 g O₂
