What amount (in mol) and what mass (in g) of water can be produced when 82.1 ug of ethane reacts with 61.4 ug of molecular oxygen? C,H, (g) +O2(8) → H,O(1) + CO,(g) -5 A: 1.64 × 10°; 2.96 × 10

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What amount (in mol) and what mass (in g) of water can be produced when 82.1 ug of ethane reacts
with 61.4 ug of molecular oxygen?
C,H, (8) +O2(g)
– H,O(1) +CO2(g)
A: 1.64 × 10°; 2.96 × 10¬5
Transcribed Image Text:What amount (in mol) and what mass (in g) of water can be produced when 82.1 ug of ethane reacts with 61.4 ug of molecular oxygen? C,H, (8) +O2(g) – H,O(1) +CO2(g) A: 1.64 × 10°; 2.96 × 10¬5
Expert Solution
Step 1

Given data :

Mass of ethane = 82.1 μg = 8.21 x 10-5 gms                 Its molar mass = 30.07 gm/mol

Mass of oxygen = 61.4 μg = 6.14 x 10-5 gms                 Its molar mass = 32 gm/mol

So,

Number of moles of ethane=Given massMolar mass=8.21×10-530.07=2.73×10-6 molesNumber of moles of O2=Given massMolar mass=6.14×10-532=1.92×10-6 moles

So, O2 is the limiting reagent here and ethane is in excess. We will use the number of moles of O2 for further calculation.

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