...when 3.35 g of hydrogen is reacted with 50.75 g of iodine to produce hydrogen iodide (HI).
H2(g) + I2(g) → 2HI(g)
For the chemical reaction below, determine the amount of HI produced...?
2016-10-03 1:51 am
回答 (1)
2016-10-03 2:05 am
Molar mass of H₂ = 1.008 × 2 g/mol = 2.016 g/mol
Molar mass of I₂ = 126.9 × 2 g/mol = 253.8 g/mol
Molar mass of HI = (1.008 + 126.9) g/mol = 127.9 g/mol
H₂(g) + I₂(g) → 2HI
Mole ratio H₂ : I₂ : HI = 1 : 1 : 2
Initial number of moles of H₂ = (3.35 g) / (2.016 g/mol) = 1.662 mol
Initial number of moles of I₂ = (50.75 g) / (253.8 g/mol) = 0.2000 mol < 1.662 mol
Hence, I₂ is the limiting reactant (limiting reagent).
Number of moles of I₂ reacted = 0.2000 mol
Number of moles of HI reacted = (0.2000 mol) × 2 = 0.4000 mol
Mass of HI reacted = (127.9 g/mol) × (0.4000 mol) = 51.16 g
Molar mass of I₂ = 126.9 × 2 g/mol = 253.8 g/mol
Molar mass of HI = (1.008 + 126.9) g/mol = 127.9 g/mol
H₂(g) + I₂(g) → 2HI
Mole ratio H₂ : I₂ : HI = 1 : 1 : 2
Initial number of moles of H₂ = (3.35 g) / (2.016 g/mol) = 1.662 mol
Initial number of moles of I₂ = (50.75 g) / (253.8 g/mol) = 0.2000 mol < 1.662 mol
Hence, I₂ is the limiting reactant (limiting reagent).
Number of moles of I₂ reacted = 0.2000 mol
Number of moles of HI reacted = (0.2000 mol) × 2 = 0.4000 mol
Mass of HI reacted = (127.9 g/mol) × (0.4000 mol) = 51.16 g
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