A solution is prepared by dissolving 2.61 g ethanol (C2H5OH) in 16.2 g water. What is the molality (mol/kg) of C2H5OH in the sol
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Answer:
Fe =52.2%, O = 44.9%, H = 2.81%
Explanation:
Percentage composition is also known as percentage by mass.
First, find the Mr of the compound.
Mr of Fe(OH)₃ = 55.8 + (16 × 3) + (1 × 3)
= 106.8
Now Divide the mass of the element, as per the compound, by the Mr of the compound found and times that by 100.
% composition of Fe = × 100 = 52.2%
% composition of O = × 100 = 44.9%
[There are 3 oxygen atoms in the compound Fe(OH)₃, so we will multiply the atomic mass of oxygen with the number of atoms in the compound: 16×3 ]
% composition of H = × 100 = 2.81%
[There are 3 hydrogen atoms in the compound Fe(OH)₃, so we will multiply the atomic mass of hydrogen with the number of atoms in the compound: 1×3 ]
8.24g
Explanation:
Given parameters;
Mass of NH₃ = 10g
Unknown:
Mass of N₂ = ?
Solution:
Reaction equation:
N₂ + 3H₂ → 2NH₃
Step 1: Convert 10.0 g of NH₃ into moles of NH₃
number of moles =
molar mass of NH₃ = 14 + 3(1) = 17g/mol
Number of moles = = 0.59mole
Step 2: Convert the moles of NH₃ calculated in Step 1 into moles of N₂
1 mole of N₂ produced 2 mole of NH₃
0.59 moles of ammonia would be formed by mole = 0.29mole of N₂
Step 3: Convert the moles of N₂ calculated in Step 2 into grams of N₂.
Mass of N₂ = number of moles x molar mass
Molar mass of N₂ = 14 x 2 = 28g/mol
Mass of N₂ = 0.29 x 28 = 8.24g
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Number of moles brainly.com/question/1841136
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