Question

Solid sodium reacts with water to produce aqueous sodium hydroxide (NaOH) and molecular hydrogen gas. Determine the limiting reactant if 8.21 grams of sodium are reacted with 6.98 grams of water.

Answer 1

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

Molar mass of Na= 23 gm

Mass of Na= 8.21 gm

number of moles of Na = mass /molar mass

                                    = 8.21/23

                                  =0.357 mol

Molar mass of water = 18 gm

mass of water = 6.98 gm

number of moles of water = mass / molar mass

                                        = 6.98/18

                                       =0.388 mol

From reaction given 2 mol of Na requires 2 mol of water

Na available is 0.357 mol whereas water available is 0.388 mol

Clearly Na (sodium) is limiting reagent