Question

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

Answer 1

Answer: 0.064 grams is the answer

Step-by-step explanation:

at the end you must multiply 88.2/100*0.073=0.064

Answer 2

Mass of hydrogen produced is 0.07g.

Step-by-step 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