How many grams of FeBr 3 would be present in 150.0 mL of a 0.435 M FeBr 3 solution?

mols = M x L = ?

Then mols = grams/molar mass. You know mols and molar mass, solve for grams.

To find the number of grams of FeBr3 present in the solution, we can use the formula:

grams = moles * molar mass

First, let's find the number of moles of FeBr3 in the solution. To do this, we will use the molarity (M) and the volume (in liters) of the solution:

moles = M * volume

Given:
Molarity (M) = 0.435 M
Volume (V) = 150.0 mL = 0.150 L

moles = 0.435 M * 0.150 L
moles = 0.06525 moles

Next, let's find the molar mass of FeBr3 (Iron (III) Bromide). The molar mass of Fe is 55.845 g/mol, and the molar mass of Br is 79.904 g/mol. Since FeBr3 has one Fe atom and three Br atoms, the molar mass can be calculated as:

molar mass = (mass of Fe) + (3 * mass of Br)

molar mass = (55.845 g/mol) + (3 * 79.904 g/mol)
molar mass = 55.845 g/mol + 239.712 g/mol
molar mass = 295.557 g/mol

Finally, we can calculate the number of grams of FeBr3 in the solution:

grams = moles * molar mass
grams = 0.06525 moles * 295.557 g/mol
grams ≈ 19.29 grams

Therefore, there would be approximately 19.29 grams of FeBr3 present in 150.0 mL of a 0.435 M FeBr3 solution.