How many milliters of 0.500 M magnesium nitrate can be produced from the reaction of 35.5 grams of magnesium bromide.

moles MgBr2 = grams/molar mass = ??

1 mole MgBr2 = 1 mole Mg(NO3)2
M Mg(NO3)2 = moles/L soln
Solve for L and convert to mL.

To determine the number of milliliters of 0.500 M magnesium nitrate that can be produced from the reaction of 35.5 grams of magnesium bromide, we need to follow these steps:

1. Write out the balanced chemical equation for the reaction between magnesium bromide and magnesium nitrate.

2. Calculate the number of moles of magnesium bromide using its molar mass.

3. Use stoichiometry to determine the mole ratio between magnesium bromide and magnesium nitrate.

4. Calculate the theoretical yield of magnesium nitrate in moles.

5. Convert the moles of magnesium nitrate to milliliters using the molarity of the solution.

Let's follow these steps one by one:

1. The balanced chemical equation for the reaction between magnesium bromide (MgBr2) and magnesium nitrate (Mg(NO3)2) is:

2 MgBr2 + Mg(NO3)2 -> 3 Mg(NO3)2 + 2 Br2

2. Calculate the number of moles of magnesium bromide:
The molar mass of magnesium bromide (MgBr2) is calculated by adding the atomic masses of magnesium (24.31 g/mol) and twice the atomic mass of bromine (79.90 g/mol) together:
Molar mass of MgBr2 = 24.31 g/mol + 2 * 79.90 g/mol = 184.11 g/mol

To convert grams to moles, use the formula:
moles = mass / molar mass

moles of MgBr2 = 35.5 g / 184.11 g/mol

3. Use stoichiometry to determine the mole ratio between magnesium bromide and magnesium nitrate:
From the balanced equation, the mole ratio of magnesium bromide to magnesium nitrate is 2:3.

4. Calculate the theoretical yield of magnesium nitrate in moles:
By using the mole ratio from step 3, we can calculate the moles of magnesium nitrate:
moles of Mg(NO3)2 = (moles of MgBr2) * (3 moles of Mg(NO3)2 / 2 moles of MgBr2)

5. Convert the moles of magnesium nitrate to milliliters using the molarity of the solution:
To convert moles to milliliters, we need to multiply by the volume (in liters) and the concentration (in moles per liter):
volume (in liters) = moles / molarity

volume (in milliliters) = (moles of Mg(NO3)2) * (0.500 mol/L) * (1000 mL/L)

Now, substitute the values obtained in each step and calculate the final answer to determine how many milliliters of 0.500 M magnesium nitrate can be produced from 35.5 grams of magnesium bromide.