If 20.0g Mg react with excess HCl to form magnesium chloride and hydrogen gas, how much MgCl 2 is produced? __Mg + __HCl → __MgCl2 + __H2

Question

Stoichiometry, chemistry, measurement

Answer 1

Start by balancing the chemical reaction equation:

Mg + 2HCl → MgCl₂ + H₂

Remember this chemical reaction equation is in units of moles, 1 mole of Mg combined with 2 moles of HCl will yield 1 mole of MgCl₂ and one mole of H₂

So to use the balanced chemical reaction equation, we must first convert the 20.0 g of Mg into moles of Mg. To do that, we go to the Periodic Table to get the molar mass of Mg. It is 24.305 g/mole.

So, # of moles of Mg = (20.0 g)/(24.305 g/mole) = 0.8229 moles. Since it takes 1 mole of Mg to make 1 mole of MgCl₂ then there is 0.8229 moles of MgCl₂ produced. Since the question doesn't specify whether they are looking for moles of MgCl₂ or mass of MgCl₂, I'm going to guess they want mass.

Since we know there are 0.8229 moles of MgCl₂ produced, we can calculate mass by again using the molar mass to convert.

Calculate molar mass of MgCl₂:

Mg: 24.305 x 1 = 24.305 g/mole

Cl: 35.453 x 2 = 70.906 g/mole

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MgCl₂ = 95.211 g/mole

Mass of MgCl₂ = (0.8229 moles of MgCl₂)(95.211 g/mole of MgCl₂) = 78.3 g MgCl₂