更新1:
NEED HELP ASAP!!!
How many molecules (not moles) of NH3 are produced from 6.19×10−4 g of H2?
回答 (2)
2017-02-03 3:11 pm
Molar mass of H₂ = (1.008 × 2) g/mol = 2.016 g/mol
Avogadro constant = 6.022 × 10²³ / mol
N₂(g) + 3H₂(g) → 2NH₃(g)
OR: Mole ratio H₂ : NH₃ = 3 : 2
No. of moles of H₂ = (6.19 × 10⁻⁴ g) / (2.016 g/mol) = 3.07 × 10⁻⁴ mol
No. of NH₃ molecules = (3.07 × 10⁻⁴ mol) × (2/3) × (6.022 × 10²³ / mol) = 1.23 × 10²⁰
Avogadro constant = 6.022 × 10²³ / mol
N₂(g) + 3H₂(g) → 2NH₃(g)
OR: Mole ratio H₂ : NH₃ = 3 : 2
No. of moles of H₂ = (6.19 × 10⁻⁴ g) / (2.016 g/mol) = 3.07 × 10⁻⁴ mol
No. of NH₃ molecules = (3.07 × 10⁻⁴ mol) × (2/3) × (6.022 × 10²³ / mol) = 1.23 × 10²⁰
2017-02-03 2:30 pm
Write the balanced equation
Use molar mass to convert gram of H2 to moles of H2
Use mole ratio (from balanced equation) to determine moles of NH3
Multiply that answer by Avogadro's number.
Use molar mass to convert gram of H2 to moles of H2
Use mole ratio (from balanced equation) to determine moles of NH3
Multiply that answer by Avogadro's number.
收錄日期: 2021-04-18 15:59:29
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