Final answer:
The volume of 2.00 M silver nitrate solution required to completely react with 5.20 g of magnesium is 0.214 L or 214 mL. Therefore, the correct option is 0.400 L since it is the smallest available volume that is greater than the required 214 mL.
Step-by-step explanation:
To determine the volume of solution needed to completely react with 5.20 g of magnesium metal when using either potassium nitrate or silver nitrate solution, first, we need to write down the relevant chemical equations for the reactions.
For Magnesium and Silver Nitrate Reaction:
Mg(s) + 2AgNO3(aq) → Mg(NO3)2(aq) + 2Ag(s)
Calculating moles of Magnesium:
Molar mass of Mg = 24.31 g/mol
Moles of Mg = mass of Mg / molar mass of Mg
Moles of Mg = 5.20 g / 24.31 g/mol ≈ 0.214 moles of Mg
Stoichiometry of Reaction:
From the equation, 1 mole of Mg reacts with 2 moles of AgNO3.
Therefore, 0.214 moles of Mg would react with 0.428 moles of AgNO3.
Calculating Volume of Silver Nitrate Solution:
Molarity of AgNO3 = 2.00 M
Volume = moles of AgNO3 / molarity of AgNO3
Volume = 0.428 moles / 2.00 M ≈ 0.214 L or 214 mL
Since only 0.800 L of AgNO3 is available, and we only need 0.214 L to completely react with the magnesium, we conclude that the 0.200 L option is not enough, and 0.400 L is more than sufficient.
Potassium Nitrate Reaction:
Potassium nitrate will not react with magnesium in this context, so it is irrelevant to the question.