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If the reaction yield is 88.2%,what mass in grams of hydrogen is produced by the reaction of 7.73 g of magnesium with 1.31 g of water
Mg + 2h2o >>mg(oh)2+h2

Respuesta :

Neetoo

Answer:

Mass of hydrogen produced is 0.07g.

Explanation:

Given data:

Mass of magnesium = 7.73 g

Mass of water = 1.31 g

Reaction yield = 88.2%

Mass of hydrogen produced = ?

Solution:

Chemical equation:

Mg + 2H₂O     →   Mg(OH)₂ + H₂

Number of moles of water:

Number of moles = mass/ molar mass

Number of moles = 1.31 g / 18 g/mol

Number of moles = 0.07 mol

Number of moles of magnesium:

Number of moles = mass/ molar mass

Number of moles = 7.73 g / 18 g/mol

Number of moles = 0.43 mol

Now we will compare the moles of water and magnesium with hydrogen.

                    Mg            :           H₂

                     1               :            1

                    0.43          :          0.43

                    H₂O          :             H₂

                      2             :              1

                   0.07           :            1/2×0.07 = 0.035

Mass of hydrogen:

Mass = number of moles × molar mass

Mass = 0.035 mol × 2 g/mol

Mass = 0.07  g

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