How many grams of Mg are needed to react completely with 3.80 L of a 2.50 M HF solution?

You need a balanced chemical equation Mg (s) + 2HF (aq) - - > MgF2 (s) + H2 (g)

3.80L x (2.50 mol HF / 1L) x (1 mol Mg / 2 mol HF) x (24.3g Mg / 1 mol Mg) = 115gMg ... To three significant digits