What volume of hydrogen is formed when 3.00 g of magnesium react with an excess of dilute sulfuric acid is
carried out under room temperature and pressure at 1 atm.

Volume of hydrogen gas formed = 3.1 L

Explanation:

Given data:

Mass of magnesium = 3.00 g

Temperature = room = 27°C (27+273 = 300 K)

Pressure = 1 atm

Volume of hydrogen gas formed = ?

Solution:

Chemical equation:

Mg + H₂SO₄    →       MgSO₄ + H₂

Number of moles of Mg:

Number of moles = mass/molar mass

Number of moles = 3.00 g/ 24 g/mol

Number of moles = 0.125 mol

Now we will compare the moles Mg with hydrogen.

                    Mg             :             H₂

                     1                :              1

                   0.125          :          0.125

Volume of hydrogen:

PV = nRT

1 atm × V = 0.125 mol  ×  0.0821 atm.L /K.mol × 300 k

V = 3.1atm.L/1 atm

V = 3.1 L

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