posted by Arlene on .
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.