Sodium metal and water react to form hydrogen and sodium hydroxide. If 11.96 g of sodium react with water to form 0.52 g of hydrogen and 20.80 g of sodium hydroxide, what mass of water was involved in the reaction

9.36g of H2O.

Explanation:

We'll begin by writing the balanced equation for the reaction.

2Na + 2H2O —> 2NaOH + H2

Next, we shall determine the mass of Na and H2O that reacted from the balanced equation.

This is illustrated below:

Molar mass of Na = 23g/mol

Mass of Na from the balanced equation = 2 x 23 = 46g

Molar mass of H2O = (2x1) + 16 = 18g/mol

Mass of H2O from the balanced equation = 2 x 18 = 36g

From the balanced equation above,

46g of Na reacted with 36g of H2O.

Now, we can determine the mass of H2O involved in reaction as follow:

From the balanced equation above,

46g of Na reacted with 36g of H2O.

Therefore, 11.96g of Na will react with = (11.96 x 36)/46 = 9.36g of H2O.

Therefore, 9.36g of H2O were used for the reaction.

they both have a y in them

i believe it is c.) 1575 j.