What volumeof hydrogen at stp by the reaction of 67.3

Complete Question:

                                   What volume of hydrogen will be produced at STP by the reaction 67.3 g of magnesium with excess water according to the following reaction?

                               Mg + 2H₂O --> Mg(OH)₂ + H₂

               62 L

Solution:

Step 1: Calculate Moles of Mg as;

                   Moles  =  Mass / M.Mass

                   Moles  =  67.3 g / 24.30 g/mol

                   Moles  =  2.76 moles of Mg

Step 2: Calculate Moles of H₂ as;

According to balance chemical equation,

                   1 mole of Mg produced  =  1 mole of H₂

So,

             2.76 moles of Mg will produce  = X moles of H₂

Solving for X,

                      X =  2.76 mol × 1 mol / 1 mol

                      X =  2.76 mol

Step 3: Calculating volume of H₂,

             1 mole of ideal H₂ occupies  =  22.4 L Volume at STP

So,

                   2.76 moles of H₂ will occupy  =  X L of H₂ at STP

Solving for X,

                     X =  2.76 mol × 22.4 L / 1 mol

                    X =  61.82 L ≈ 62 L

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