Final answer:
To calculate the grams of AgCl produced, we must use stoichiometry and convert volume to moles using molarity. The correct calculation for the mass of AgCl exceeds the provided answer options, suggesting a potential issue with the question's setup or answer choices.
Step-by-step explanation:
To calculate the grams of silver chloride produced from 10.00 ml of 10.0M magnesium chloride with 100.0 ml of 2.20 M silver nitrate, we must first convert volume to moles for each reactant using their molarities.
MgCl2: 10.00 ml × 10.0M = 0.1000 moles
AgNO3: 100.0 ml × 2.20 M = 0.220 moles
The balanced chemical equation is:
2AgNO
3
+ MgCl
2
→ Mg(NO
3
)
2
(s) + 2AgCl(aq)
Since silver nitrate is the limiting reactant (as we need 2 moles of AgNO3 for every mole of MgCl2), we calculate the moles of AgCl using the stoichiometry of the reaction:
Moles of AgCl produced = 0.220 moles AgNO3 × (2 moles AgCl / 2 moles AgNO3) = 0.220 moles AgCl
To find the mass of AgCl, we use the molar mass of AgCl (143.32 g/mol):
Mass of AgCl = 0.220 moles × 143.32 g/mol = 31.53 g
However, this calculation doesn't take into account the amount of MgCl2 that would have capped the AgCl production, as it would have run out first; therefore, using 0.100 moles of MgCl2, the correct calculation should be:
Mass of AgCl = 0.1000 moles MgCl2 × (2 moles AgCl / 1 mole MgCl2) × 143.32 g/mol = 28.66 g
However, the answer options given (A) 5.00 g, B) 10.0 g, C) 15.0 g, D) 20.0 g) do not match our calculated mass, indicating either a typo or miscalculation in the provided options or the original problem setup. In this case, we should consult with the provider of the question to clarify the parameters or what the correct answer choices should be.