Sodium metal and water react to form hydrogen and sodium hydroxide. If 1.99 g of sodium react with water to form 0.087 g of hydrogen and 3.47 g of sodium hydroxide, what mass of water was involved in the reaction?

1.56 g of water was involved in the reaction

Explanation:

From the stoichiometric equation

2Na + 2H2O = 2NaOH + H2

NB : Mm Na= 23, Mm H2O = ( 2+16)= 18

2(23) of Na requires 2(18) of water

Hence 1.99g of Na will require 1.99×2×18/2(23) of water = 1.56 g of water

1.6 g.

Explanation:

On of the laboratory process for the preparation of Hydrogen is the reaction of sodium with water. The balanced equation for the reaction is given below;

2Na(s) + 2H2O(g) > 2NaOH(aq) + H2(g).

From the equation of reaction above we have 2 moles of Na reacting with 2 moles of water to give 2 moles of sodium Hydroxide, NaOH and one mole of Hydrogen.

That is;

2 : 2 > 2 : 1.

So, the molar mass of NaOH = 40 grams per mole.

And from the question we are given the following parameters; mass of NaOH = 3.47 g, mass of Hydrogen formed= 0.087 g , mass of Sodium, Na = 1.99 g.

Hence, the mass of water involved in the reaction =[ (2 × 18) / 80 ] × 3.47 grams.

mass of water involved in the reaction = 0.45 × 3.47 grams.

mass of water involved in the reaction = 1.5615 g.

93.2 ml + 3920 ml = 4013 ml total  

(93.2 ml) * (2.03 m) / (4013 ml) = 0.0471 m

the volume in in l of the diluted solution would be 0.0471 m.