Answer:
Moles of silver iodide produced = 1.4 mol
Explanation:
Given data:
Mass of calcium iodide = 205 g
Moles of silver iodide produced = ?
Solution:
Chemical equation:
CaI₂ + 2AgNO₃ → 2AgI + Ca(NO₃)₂
Number of moles calcium iodide:
Number of moles = mass/ molar mass
Number of moles = 205 g/ 293.887 g/mol
Number of moles = 0.7 mol
Now we will compare the moles of calcium iodide with silver iodide.
CaI₂ : AgI
1 : 2
0.7 : 2×0.7 = 1.4
Thus 1.4 moles of silver iodide will be formed from 205 g of calcium iodide.
The electron group arrangement of NO²⁻is trigonal planar. The molecular shape is bent, and the bond angle is 120°.
<h3>What is the molecular shape of a compound?</h3>
The molecular geometry of the compound shows the position of nuclei and the electron of the compound. It shows how the joining of electrons and nuclei makes the shape of the compound.
Like here, the shape of nitrite is bent with lone pair which is shown by Lewis's structure The bond angle will be the distance between the nuclei of the neighbor atoms.
Thus, the electron geometry arrangement of nitrite is trigonal planer with a bent shape and the bond angle will be 120°.
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The valence electron does the halogens possess are 7
- Valence electrons are found in the outermost energy level of an atom
- They are involved in the formation of chemical bonding with other atoms.
- The halogens elements are found in group 17 on the periodic table
- The halogens include fluorine, chlorine, bromine, iodine and astatine.
- They have seven valence electrons, so they are extremely reactive as they only need one more to fill their outer shell.
- By octet rule we can say that the electron with 8 outer most shell is full and stable.
Hence the halogens posses 7 valence electron
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Answer:
The answer to your question is
1.-Fe₂O₃
2.- 280 g
3.- 330 g
Explanation:
Data
mass of CO = 224 g
mass of Fe₂O₃ = 400 g
mass of Fe = ?
mass of CO₂
Balanced chemical reaction
Fe₂O₃ + 3CO ⇒ 2Fe + 3CO₂
1.- Calculate the molar mass of Fe₂O₃ and CO
Fe₂O₃ = (56 x 2) + (16 x 3) = 160 g
CO = 12 + 16 = 28 g
2.- Calculate the proportions
theoretical proportion Fe₂O₃ /3CO = 160/84 = 1.90
experimental proportion Fe₂O₃ / CO = 400/224 = 1.78
As the experimental proportion is lower than the theoretical, we conclude that the Fe₂O₃ is the limiting reactant.
3.- 160 g of Fe₂O₃ --------------- 2(56) g of Fe
400 g of Fe₂O₃ --------------- x
x = (400 x 112) / 160
x = 280 g of Fe
4.- 160 g of Fe₂O₃ --------------- 3(44) g of CO₂
400 g of Fe₂O₃ -------------- x
x = (400 x 132)/160
x = 330 gr
