Sodium Chloride is ordinary table salt
You have a small sodium atom with a positive charge Na+, and a larger chlorine atom with a negative charge, making it a Chloride ion, Cl-.
This ion formation has resulted from Na transferring one of its electrons to Cl. there opposite charges then attract them to each-other.
Because the electrons are transferred rather than shared, we know that the bond is Ionic, rather than covalent.
Most ionic compounds, if not all, are salts that form a crystal lattice structure, due to the opposite charges in the molecule.
Think of it like this
Na+—Cl-
Cl—-Na+
Na+—Cl-
That’s how the molecules attract and stick to each other. That would continue until you ran out of NaCl molecules.
Answer:
84.4g of AgCl
Explanation:
Based on the reaction:
2AgNO₃ + CaCl₂ → 2AgCl + Ca(NO₃)₂
<em>2 moles of AgNO₃ and 1 mole of CaCl₂ priduce 2 moles of AgCl and 1 mole of Ca(NO₃)₂</em>
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100g of each reactant are:
AgNO₃: 100g × (1mol / 169.87g) = 0.589 moles
CaCl₂: 100g × (1mol / 110.98g) = 0.901 moles
For a complete reaction of 0.901 moles of CaCl₂ are necessaries 0.901×2 = <em>1.802 moles of AgNO₃. </em>As there are just 0.589moles, <em>AgNO₃ is limitng reactant</em>
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0.589 moles of AgNO₃ produce:
0.589 moles × ( 2 moles AgCl / 2 moles AgNO₃) =
<em>0.589 moles of AgC</em>l. In mass:
0.589 moles of AgCl × (143.32g / mol) =<em> 84.4g of AgCl</em>
The mass of 
that would be formed will be 18.22 grams
<h3>Stoichiometric calculations</h3>
Let us first look at the balanced equation of the reaction:
The mole ratio of Y to 
is 2:3.
Mole of 10.0 grams of Y = 10/88.9 = 0.11 moles
Mole of 10.0 grams = 10/71 = 0.14 moles
3/2 of 0.11 = 0.165. Thus, is limiting in availability.
Mole ratio of and = 3:2
Equivalent mole of = 2/3 x 0.14 = 0.093 moles.
Mass of 0.093 moles =0.093 x 195.26 = 18.22 grams
More on stoichiometric calculations can be found here: brainly.com/question/27287858
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