# In the reaction of interaction of sodium weighing 2.3 g with an excess of water, sodium hydroxide is formed.

In the reaction of interaction of sodium weighing 2.3 g with an excess of water, sodium hydroxide is formed. Calculate the number of moles of hydrochloric acid that is needed to neutralize sodium hydroxide.

Let’s write down the reaction equations.
2Na + 2HOH = 2NaOH + H2.
Let us determine the mass of sodium hydroxide by the reaction equation.
M (Na) = 23 g / mol.
M (NaOH) = 23 + 16 + 1 = 40 g / mol.
Let’s make a proportion.
2.3 g Na – x g NaOH.
2 * 23 g / mol – 2 * 40 g / mol.
X = 4 g.
Next, we write down the reaction equation.
NaOH + HCl = NaCl + H2O.
Determine the amount of sodium hydroxide substance.
n = m / M.
n (NaOH) = 4/40 = 0.1 mol.
Let us determine the amount of hydrochloric acid substance according to the reaction equation.
0.1 mol Na – y mol HCl.
1 mol – 1 mol.
Y = 0.1 mol HCl.
Answer: n (HCl) = 0.1 mol.

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