# Limiting reactant and percent yield [closed]

If 2.30 mol of sodium reacts with 1.95 mol of water, how many moles of sodium hydroxide are produced?

Would it be 2.3 mol NaOH?

The first step is to figure out a reaction equation.

[I'm assuming elemental] sodium reacts with water violently because it releases a lot of energy when its only valence electron gets ripped apart. It creates \ce{NaOH} and \ce{H2}.

$\ce{Na + H2O -> NaOH + H2}$

We need to balance this.

$\ce{2Na + 2H2O -> 2NaOH + H2}$

$\ce{Na}$ and $\ce{H2O}$ are consumed at equal rates, so the $1.95$ moles of $\ce{H2O}$ are going to be used up completely (leaving $2.30 - 1.95 = 0.35$ moles of $\ce{Na}$ not reacted).

$\ce{NaOH}$ is formed at the same rate of consumption of the other two, so it's going to have $1.95$ moles produced.

--

Mathematically...

$\frac{2.30mol_\ce{Na}}{2mol_\ce{Na}} = 1.15$

and

$\frac{1.95mol_\ce{water}}{2mol_\ce{water}} = 0.975$

Effectively, water burns out faster in the reaction than sodium. So, we have to base our calculations off of moles of water.

For every 2 moles of water used, 2 moles of sodium hydroxide are formed.

$1.95mol_\ce{water} * \frac{2mol_\ce{NaOH}}{2mol_\ce{water}} = 1.95mol_\ce{NaOH}$

$$\ce{Na + H2O->NaOH + \frac12H2}$$

If 2.30 mol of sodium reacts with 1.95 mol of water, how many moles of sodium hydroxide are produced? Would it be 2.3 mol NaOH?

No cause the limiting reagent here is water which is in lesser amount, since both reactants have same stoichiometric coefficients, so the product will be formed in accordance with amount of water, i.e. 1.95 moles of $\ce{NaOH}$