Chemistry Help Please?

[匿名]

2016-02-29 7:56 pm

更新1:

When 52.8 g of calcium and 30.4 g of nitrogen gas undergo a reaction that has a 88.0% yield, what mass (g) of calcium nitride forms?

回答 (1)

賣女孩的火柴

2016-02-29 8:25 pm

✔ 最佳答案

Molar mass of Ca = 40 g/mol
Molar mass of N₂ = 14 × 2 = 28 g/mol
Molar mass of Ca₂N₃ = 40×2 + 14×3 = 122 g/mol

4Ca + 3N₂ → 2Ca₂N₃
Mole ratio Ca : N₂ : Ca₂N₃ = 4 : 3 : 2

Initial no. of moles of Ca = 52.8/40 = 1.32 mol
Initial no. of moles of N₂ = 30.4/28.0 = 1.086 mol

When Ca completely reacts :
No. of moles of N₂ needed = 1.32 × (3/4) = 0.99 mol < 1.086 mol
Thus, N₂ is in excess, and Ca is the limiting reactant.

No. of moles of Ca reacted = 1.32
Theoretically, no. of moles of Ca₂N₃ formed = 1.32 × (2/4) = 0.66 mol
Actual mass of Ca₂N₃ formed = 0.66 × 122.2 × 88.0% = 70.9 g

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