# How much water is formed when 1.0 mol of HCl reacts completely with 1.0 mol of NaOH?

May 13, 2016

Assume that:
n = number of moles
m = mass of substance
M = molar mass

Looking at the equation:
$H C l + N a O H \implies {H}_{2} O + N a C l$.

Since all compounds in this reaction have 1 mole, it will not affect the number of moles (n).

If 1 mole of HCl or NaOH gives you 1 mole of ${H}_{2} O$, then the number of moles in ${H}_{2} O$ is: [$1 \div 1 \times 1$] = 1 mole.

Now you have to find out the molar mass (M) of water (${H}_{2} O$).
Before that, take note that you know what atoms are present in ${H}_{2} O$ - in that case, hydrogen and oxygens are present.

Refer to your periodic table and you can see molar mass of (?) is:
Hydrogen = 1.0 g/mol
Oxygen = 16.0 g/mol

Therefore the molar mass (M) of ${H}_{2} O$ is:
[$2 \times 1.0 + 1 \times 16.0$] = 18.0 g/mol.

Last step, find the mass (m) of water.
The formula of finding mass of substance (m) is:
$m = n \times M$.
The mass of ${H}_{2} O$ is: [m = 1.0 moles $\times$ 18.0 g/mol = 18.0 grams]

Therefore 18.0 grams of water is formed when 1 mole of HCl reacts completely with 1 mole of NaOH.