A piece of metallic sodium weighing 36.8 g was dropped into 1 liter of water.

A piece of metallic sodium weighing 36.8 g was dropped into 1 liter of water. Calculate the mass fraction of the obtained substance in water.

To solve the problem, we will compose the reaction equation, arrange the coefficients:
2Na + 2H2O = 2NaOH + H2 – redox reaction, sodium hydroxide and hydrogen are released;
Determine the molecular weights of the compounds:
M (Na) = 22.9 g / mol;
M (H2O) = 18 g / mol;
M (NaOH) = 22.9 + 1 + 16 = 39.9 g / mol;
Let’s calculate the amount of sodium moles by the formula:
Y (Na) = m / M = 36.8 / 22.9 = 1.6 mol;
Let’s make a proportion, determine the amount of moles of water:
1.6 mol (Na) – X mol (H2O);
-2 mol -2 mol hence, X mol (H2O) = 1.6 * 2/2 = 1.6 mol;
Calculations show that the amounts of reagents are equal.
Let’s calculate the number of moles of the reaction product – sodium hydroxide:
1.6 mol (Na) – X mol (NaOH);
2 mol -2 mol hence, X mol (NaOH) = 1.6 * 2/2 = 1.6 mol;
Let us determine the theoretical mass of NaOH:
m (theoretical) = Y * M = 1.6 * 39.9 = 63.8 g;
Let’s calculate the practical mass, taking into account that the amounts of substances are proportionally equal:
m (practical) = Y * M = 1.6 * 39.9 = 63.84 g;
The reaction product yield is:
W = m (practical) / m (theoretical) * 100 = 63.8 / 63.84 * 100 = 99.9%;
Answer: the yield of the reaction product of sodium hydroxide is 99.9%.