Answer:
Net ionic equation: Ag⁺(aq) + Cl⁻(aq) → AgCl(s)
44.9% as AgNO₃
Explanation:
When sodium nitrate, NaNO₃ and silver nitrate, AgNO₃ are dissolved in water, the Na⁺, NO₃⁻ and Ag⁺ ions are formed.
Then, the addition of NaCl (Na⁺ and Cl⁻) produce AgCl⁻ as precipitate. <em>The net ionic equation is:</em>
<h3>Ag⁺(aq) + Cl⁻(aq) → AgCl(s)</h3><h3 />
If 2.54g of AgCl are formed and represents the 95.9% of yield. The real amount of AgCl is:
2.54g AgCl * (100% / 95.9%) = 2.65g AgCl.
In moles (Molar mass AgCl = 143.32g/mol):
2.65g AgCl * (1mol / 143.32g) = 0.0185 moles of AgCl = Moles of AgNO₃
<em>Because all Ag comes from AgNO₃</em>
<em />
Thus, the original mass of silver nitrate and its precentage is (Molar mass AgNO₃ = 169.87g/mol):
0.0185 moles AgNO₃ * (169.87g / mol) = 3.14g of AgNO₃
Percentage:
3.14g AgNO₃ / 7.00g * 100 =
<h3>44.9% as AgNO₃</h3>
