A<span>toms are always bonded so that nearby atoms have to form into a compound and can be classified into different forms of compounds or molecules. It is also integral in our everyday life.</span>
You need to calculate the molar mass for Al(CN)3 using the atomic weights for Al, C, and N given on the periodic table.
1 Al (26.98) + 3 C (3 x 12.01) + 3 N (3 x 14.01) = 105.04 g Al(CN)3 / mole
183 g Al(CN)3 x (1 mole Al(CN)3 / 105.04 g Al(CN)3) = 1.74 moles Al(CN)3
Answer:
Percent yield = 57%
Explanation:
Given data:
Mass of nitrogen = 15.0 g
Mass of hydrogen = 15.0 g
Mass of ammonia produced = 10.5 g
Percent yield = ?
Solution:
Chemical equation:
N₂+ 3H₂ → 2NH₃
Number of moles of hydrogen:
Number of moles = mass/molar mass
Number of moles = 15.0 g/ 2 g/mol
Number of moles = 7.5 mol
Number of moles of nitrogen:
Number of moles = mass/molar mass
Number of moles = 15.0 g/ 28 g/mol
Number of moles = 0.54 mol
Now we will compare the moles of ammonia with nitrogen and hydrogen from balance chemical equation.
N₂ : NH₃
1 : 2
0.54 : 2×0.54 = 1.08
H₂ : NH₃
3 : 2
7.5 : 2/3×7.5= 5 mol
Theoretical yield of ammonia:
Mass = number of moles × molar mass
Mass = 1.08 × 17 g/mol
Mass = 18.36 g
Percent yield:
Percent yield = (actual yield / theoretical yield) ×100
Percent yield = (10.5 g/ 18.36 g) ×100
Percent yield = 0.57 ×100
Percent yield = 57%
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