Answer:
n l m
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1 0 0 1s 1 2 2
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2 0 0 2s 1 2
2 1 1,0,-1 2p 3 6 8
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3 0 0 3s 1 2
3 1 1,0,-1 3p 3 6
3 2 2,1,0,-1,-2 3d 5 10 18
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4 0 0 4s 1 2
4 1 1,0,-1 4p 3 6
4 2 2,1,0,-1,-2 4d 5 10
4 3 3,2,1,0,-1,-2,-3 4f 7 14 32
Explanation:
This question is asking for an element with 5 valence electrons. Just go to the row it is in (excluding transition metals) and count over.
The answer would be c. P
We determine the limiting reactant by using the moles present in the equation and the actual moles.
According to equation, ratio of Fe₂O₃ : Al = 1 : 2
Actual moles of Fe₂O₃ = 187.3 / (56 x 2 + 16 x 3)
= 1.17
Actual moles of Al = 94.51 / 27
= 3.5
Fe₂O₃ is limiting. Fe₂O₃ required:
(moles Al)/2 = 3.5/2 = 1.75
Moles to be added = 1.75 - 1.17
= 0.58
Mass to be added = moles x Mr
= 0.58 x (56 x 2 + 16 x 3)
= 92.8 grams
