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

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