Question

Suppose you have 5.20 g of powdered magnesium metal, 0.800 L of 2.00 M potassium nitrate solution, and 0.800 L of 2.00 M silver nitrate solution. What volume of solution is needed to completely react with the magnesium?

Options:

0.200 L
0.400 L
0.600 L
0.800 L

Answer 1

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.