What volume of hydrogen gas at STP will be produced from 36.7 g of magnesium reacting with an excess amount of hydrochloric acid?

V = 34.3 L

Explanation:

Given data:

Volume of hydrogen produced = ?

Mass of magnesium = 36.7 g

Temperature and pressure = standard

Solution:

Mg + 2HCl       →   MgCl₂ + H₂

Number of moles of magnesium:

Number of moles = mass/molar mass

Number of moles = 36.7 g/ 24 g/mol

Number of moles = 1.53 mol

Now we will compare the moles of magnesium with hydrogen.

                         Mg           :          H₂

                          1              :           1

                         1.53          :          1.53

Number of moles of hydrogen produced = 1.53 mol

Volume of hydrogen:

PV = nRT

V = nRT / P

V = 1.53 mol × 0.0821 atm.L/mol.K × 273 K/ 1 atm

V = 34.3 L

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