One method for reclaiming silver metal from silver chloride results in a 94.6%
yield. Calculate the actual mass of silver that can be produced in this reaction
if 100.0 g of silver chloride is converted to silver metal.
2AgCl(s) y 2Ag(s) . Cl2(g)
100.0 g AgCl * 1 mol AgCl/143.5 g AgCl * 2 mol Ag/2 mol AgCl * 108 g Ag/1 mol = 75.3 g Ag, mass of Ag(s) reclaimed = 0.946 * 75.3 g Ag = 71.2 g Ag. Hope it helps :)
To calculate the actual mass of silver that can be produced in this reaction, we need to use the concept of yield. Yield is a measure of how much product is actually obtained compared to the theoretical amount that could be obtained.
In this case, the yield is given as 94.6%. This means that 94.6% of the theoretical amount of silver will be obtained.
First, we need to determine the molar mass of silver chloride (AgCl), which consists of one atom of silver (Ag) and one atom of chlorine (Cl). The molar mass of Ag is 107.87 g/mol, and the molar mass of Cl is 35.45 g/mol. So, the molar mass of AgCl is:
Molar mass of AgCl = (1 × molar mass of Ag) + (1 × molar mass of Cl)
= (1 × 107.87 g/mol) + (1 × 35.45 g/mol)
= 143.32 g/mol
Next, we calculate the theoretical amount of silver that can be obtained from 100.0 g of silver chloride using stoichiometry. The balanced equation tells us that 2 moles of Ag are produced for every 2 moles of AgCl consumed.
Moles of AgCl = Mass of AgCl / Molar mass of AgCl
= 100.0 g / 143.32 g/mol
≈ 0.698 mol
Since the reaction is 2 moles of AgCl to 2 moles of Ag, we can conclude that this will also produce 0.698 mol of Ag.
Finally, we can calculate the actual mass of silver produced using the yield:
Actual mass of Ag = Yield × Theoretical mass of Ag
= 0.946 × (0.698 mol × 107.87 g/mol)
≈ 74.72 g
Therefore, the actual mass of silver that can be produced in this reaction is approximately 74.72 g.
poop
Well, well, well! Looks like we've got a chemical conundrum on our hands! Don't worry, I'm here to clown around and provide the answer.
Since we know the yield is 94.6%, we can say that 94.6% of the silver chloride will be converted into silver metal. That means we need to multiply the initial mass of silver chloride by 0.946:
100.0 g AgCl * 0.946 = 94.6 g Ag
Voila! The actual mass of silver that can be produced in this reaction is 94.6 grams. Keep in mind, though, there might still be a little chlorine clowning around in the form of Cl2 gas.
Write the equation which you have BUT you have omitted the yield sign and and + signs. No way to know the difference between reactants and products.
Convert 100.0 g AgCl to mols.
Using the coefficients in the balanced equation, convert mols AgCl to mols Ag.
Convert mols Ag to grams Ag by grams = mols x molar mass.
Substitute in the earlier problem for the percent yield and solve for the actual amount. Post your work if you get stuck.