Ecology is the study of what
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Answer:
3.64g
Explanation:
Given parameters:
Mass of NH₃ = 18.1g
Mass of Cu₂O = 90.4g
Unknown:
Limiting reactant = ?
Mass of N₂ formed = ?
Solution:
The reaction equation is given as:
Cu₂O + 2NH₃ → 6Cu + N₂ + 3H₂O
The limiting reactant is the one in short supply in the reaction. Let us find the number of moles of the given species;
Number of moles =
Molar mass of Cu₂O = 2(63.6) + 16 = 143.2g/mol
Molar mass of NH₃ = 14 + 3(1) = 17g/mol
Number of moles of Cu₂O = = 0.13moles
Number of moles of NH₃ = = 5.32moles
From this reaction;
1 mole of Cu₂O combines with 2 mole of NH₃
So 0.13moles of Cu₂O will combine with 0.13 x 2 mole of NH₃
= 0.26moles of NH₃
Therefore, Cu₂O is the limiting reactant. Ammonia is in excess;
Mass of N₂;
Mass = number of moles x molar mass
1 mole of Cu₂O will produce 1 mole of N₂
0.13 mole of Cu₂O will produce 0.13 mole of N₂
Mass = 0.13 x (2 x 14) = 3.64g
<u>Answer:</u> The ratio of carbon in both the compounds is 1 : 2
<u>Explanation:</u>
Law of multiple proportions states that when two elements combine to form two or more compounds in more than one proportion. The mass of one element that combine with a given mass of the other element are present in the ratios of small whole number. For Example:
- <u>For Sample 1:</u>
Total mass of sample = 100 g
Mass of carbon = 27.2 g
Mass of oxygen = (100 - 27.7) = 72.8 g
To formulate the formula of the compound, we need to follow some steps:
- <u>Step 1:</u> Converting the given masses into moles.
Moles of Carbon =
Moles of Oxygen =
- <u>Step 2:</u> Calculating the mole ratio of the given elements.
For the mole ratio, we divide each value of the moles by the smallest number of moles calculated which is 2.26 moles.
For Carbon =
For Oxygen =
- <u>Step 3:</u> Taking the mole ratio as their subscripts.
The ratio of C : O = 1 : 2
Hence, the formula for sample 1 is
- <u>For Sample 2:</u>
Total mass of sample = 100 g
Mass of carbon = 42.9 g
Mass of oxygen = (100 - 42.9) = 57.1 g
To formulate the formula of the compound, we need to follow some steps:
- <u>Step 1:</u> Converting the given masses into moles.
Moles of Carbon =
Moles of Oxygen =
- <u>Step 2:</u> Calculating the mole ratio of the given elements.
For the mole ratio, we divide each value of the moles by the smallest number of moles calculated which is 3.57 moles.
For Carbon =
For Oxygen =
<u>Step 3:</u> Taking the mole ratio as their subscripts.
The ratio of C : O = 1 : 1
Hence, the formula for sample 1 is
In the given samples, we need to fix the ratio of oxygen atoms.
So, in sample one, the atom ratio of oxygen and carbon is 2 : 1.
Thus, for 1 atom of oxygen, the atoms of carbon required will be =
Now, taking the ratio of carbon atoms in both the samples, we get:
Hence, the ratio of carbon in both the compounds is 1 : 2