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What is the minimum volume of 5.0 M HCl required to completely dissolve 24.0 g of magnesium metal?
What is the minimum volume of 5.0 M HCl required to completely dissolve 24.0 g of magnesium metal?
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2 HCl + Mg → H2 + MgCl2
(24.0 g Mg) / (24.3051 g Mg/mol) x (2/1) / (5.0 mol/L HCl) = 0.39 L HCl
(24.0 g Mg) / (24.3051 g Mg/mol) x (2/1) / (5.0 mol/L HCl) = 0.39 L HCl