Question

consider the reaction between Mg(s) and HCl(aq) to producer aqueous magnesium chloride and hydrogen gas. How many liters of hydrogen gas at STP will be produced when 12.15 g of magnesium reacts with an excess of hydrochloric acid?

Answer 1

a) Let's first calculate # of moles of Mg (s) in 12.15 g of Mg.

# of Moles of Mg = Given Mass of Mg / Atomic Mass of Mg = 12.15 g / 24.30 g/moles = 0.5 moles.

A pertinent equation for the reaction between Mg (s) & HCl (aq) is,

Mg (s) + 2 HCl (aq) MgCl2 (aq) + H₂ (g).

A mole equivalence for the reaction is,

1 mole of Mg (s) 1 mole of H₂ (g).

0.5 mole of Mg (s) 0.5 mole of H₂ (g).

Theoretically, on the complete reaction of 12.15 g (i.e 0.5 mole) Mg, 0.5 (i.e. 1/2 or half) moles of H₂ (g) will be produced.

According to the Avogadro's Hypothesis, at STP, 1 mole of n ideal gas occupies 22.414 L of the volume.

Assuming H2 (g) as an Ideal gas,

0.5 moles of H2 (g) will occupy 22.414 / 2 = 11.207 L of the volume.

Volume occupied at STP is 11.207 L.

Answer: 11.207 L of hydrogen gas at STP will be produced when 12.15 g of magnesium reacts with an excess of hydrochloric acid.

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