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Answer:
Explanation:
When percentage composition is given, and asked for the empirical formula, it is simplest to assume 100 g of material. Thus,
Mass C = 40.92 g. Moles C = 40.92 g x 1 mole/12 g = 3.41 moles C
Mass H = 4.58 g. Moles H = 4.58 g x 1 mole/1.0 g = 4.58 moles H
Mass O = 54.50 g. Moles O = 54.50 g x 1 mole/16 g = 3.41 moles O
Now, we want to get the moles into whole numbers, so we begin by dividing all by the smallest, i.e. divide all values by 3.41.
Moles C = 3.41/3.41 = 1
Moles H = 4.58/3.41 = 1.34
Moles O = 3.41/3.41 = 1
Now, in order to get 1.34 to be a whole number we multiply it (and all others) by 3
Moles C = 1x3 = 3
Moles H = 1.34x3 = 4
Moles O = 1x3 = 3
Empirical Formula
Answer:
6.25 moles of N₂ is produced, and 18.8 moles of Cu and H₂O is produced.
Explanation:
We are given the chemical equation:
And we want to determine the amount of products produced when 12.5 moles of NH₃ is reacted with excess CuO.
Compute using stoichiometry. From the equation, we can see the following stoichiometric ratios:
- The ratio between NH₃ and N₂ is 2:1. (i.e. One mole of N₂ is produced from every two moles of NH₃.)
- The ratio between NH₃ and Cu is 2:3.
- The ratio between NH₃ and H₂O is 2:3. (i.e. Three moles of H₂O or Cu is produced frome every two moles of NH₃.)
Dimensional Analysis:
- The amount of N₂ produced:
- The amount of Cu produced:
- And the amount of H₂O produced:
In conclusion, 6.25 moles of N₂ is produced, and 18.8 moles of Cu and H₂O is produced.