Question

What volume of H2 gas will be produced at 1.0 atm and 28 degrees Celsius if 10g of Na are mixed with 100g of H2O.The other product of the reaction is NaOH

Answer 1

2 Na +    2 H2O -----------------------> 2 NaOH   + H2

46 g         36g                                                     2 g

10 g          100 g

46 g Na---------------------> 36 g H2O

10 g Na-------------------> 36 x 10 / 46 = 7.83 g

we need only 7.83 g H2O . but we have 100 g H2O . so it is excess reagent 'limiting reagent is Na

46 g Na ----------------------> 2 g H2

10 g Na----------------------> 10 x 2 / 46 = 0.435 g

mass of H2 formed = 0.435 g

moles of H2 formed = mass / molar mass

                                = 0.435 / 2

                                 = 0.217

Temeprature = 28 + 273 = 301 K

P V = n R T

1.0 x V = 0.217 x 0.021 x 301

V = 5.37 L

volume of H2 produced = 5.37 L