hey can someone help me out! Thanks
When 99.5 g of calcium and 45.0 g of nitrogen gas undergo a reaction that has a 76.2% yield, what mass of calcium nitride forms?
- anonymousLv 71 month ago
balanced reaction equation:
3 Ca + N₂ ⟶ Ca₃N₂
original number of moles of Ca:
(99.5 g Ca) x (1 mole Ca / 40.08 g Ca) = 2.48 mol Ca
original number of moles of N₂ :
(45.0 g N₂) x (1 mole N₂ / 28.01 g N₂) = 1.61 mol N₂
From the balanced equation, 1.61 mol N₂ reacts with 3 times as much Ca, namely, 4.83 moles Ca. There is less Ca present than that, so Ca is the limiting reactant.
Theoretical yield in grams of Ca₃N₂ from 2.48 mol Ca:
(2.48 moles Ca) x (1 mole Ca₃N₂ / 3 moles Ca) x (148.25 g Ca₃N₂ / 1 mole Ca₃N₂)
= 122.55 g Ca₃N₂
actual yield of Ca₃N₂ using percent yield information:
(122.55 g Ca₃N₂) x 0.762 = 93.4 g Ca₃N₂ …….. to 3 sig figs