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

Mg + 2HF yields MgF2 + H2

A. 4.75 g

B. 9.50 g

C. 115 g

D. 190 g

Molarity = number of moles of solute / liters of solution

Therefore,

number of moles = molarity * volume

number of moles of HF = 2.5 * 3.8 = 9.5 moles

From the balanced equation given:

one mole of Mg is required to react with two moles of HF, therefore, to get the number of moles of Mg that reacts with 9.5 moles of HF, we will just do a cross multiplication as follows:

number of moles of Mg = (9.5*1) / 2 = 4.75 moles

From the periodic table:

molar mass of Mg = 24.305 grams

number of moles = mass / molar mass

therefore,

mass = number of moles * molar mass

mass og Mg = 4.75 * 24.305 = 115.44 grams

Answer: (c)