# If "2300.0 g" of "Na"(s) reacts in a sufficient amount of water, what is the theoretical yield of "H"_2(g) in grams?

Aug 16, 2015

Sodium metal $\left(N a \left(s\right)\right)$ reacts in a single-replacement reaction, like so (I'll break it down into steps, where the middle dots mean there's a lone electron, and $\implies$ means the reaction step is taken as-written):

$1. H - O H \left(l\right) \implies H \cdot \left(g\right) + \cancel{\cdot O H \left(a q\right)}$
$2. N a \cdot \left(s\right) + \cancel{\cdot O H \left(a q\right)} \implies N a O H \left(a q\right)$

$1.$ The $H - O H$ bond is split in two. This is a violent step, and in real life there's bubbling in the water!
$2.$ $N a \cdot$ then bonds with $\cdot O H$.

Overall, we get the product of a (radical) single-replacement reaction:

$\textcolor{g r e e n}{N a \cdot \left(s\right) + {H}_{2} O \left(l\right) \to N a O H \left(a q\right) + H \cdot \left(g\right)}$

or

$\textcolor{g r e e n}{2 N a \left(s\right) + 2 {H}_{2} O \left(l\right) \to 2 N a O H \left(a q\right) + {H}_{2} \left(g\right)}$

The first obvious piece of information is that your limiting reagent is sodium metal (since the water is in excess), so you can use that as the starting compound for your "mole-bridge" $\left({\text{g"_1->"mol"_1/"g"_1->"mol"_2/"mol"_1->"g"_2/"mol}}_{2}\right)$ conversion.

Now, since we have $2300.0 \text{g}$ of Na (that's going to be rather expensive... and explosive! Don't try this at home!):

$2300.0 \cancel{g N a} \cdot \frac{\cancel{1 m o l N a}}{22.989 \cancel{g N a}} \cdot \frac{\cancel{1 m o l H}}{\cancel{1 m o l N a}} \cdot \frac{1.0079 g H}{\cancel{1 m o l H}}$

$= \textcolor{b l u e}{100.84 \text{g} H \left(g\right)}$

Aug 16, 2015

The theoretical yield of hydrogen gas is $\text{100.84 g}$.

#### Explanation:

We need to start with a balanced equation. This will give us the mole ratio between sodium metal and hydrogen gas. We will also need the molar masses of sodium and hydrogen gas in order to convert between moles and mass.

Balanced Equation
"2Na(s)"+"2H"_2"O(l")$\rightarrow$$\text{2NaOH(aq)"+"H"_2("g)}$

Mole ratio between $\text{Na}$ and $\text{H"_2}$

$\left(2 \text{mol Na")/(1"mol H"""_2}\right)$ and $\left(1 \text{mol H"""_2)/(2"mol Na}\right)$

Molar masses of sodium and hydrogen gas

$\text{Na} :$$\text{22.990 g/mol}$

${\text{H}}_{2} :$$\text{2.0159 g/mol}$

Convert $\text{2300.0 g Na}$ to moles $\text{Na}$ using its molar mass.

$2300.0 \cancel{\text{g Na"xx(1"mol Na")/(22.990cancel"g Na")="100.04 mol Na}}$

Convert moles of $\text{Na}$ to moles of ${\text{H}}_{2}$ using the mole ratio from the balanced equation .

$100.04 \cancel{\text{mol Na"xx(1"mol H"_2)/(2cancel"mol Na")="50.020 mol H"""_2}}$

Theoretical Yield of Hydrogen Gas

Convert moles of $\text{H"_2}$ to mass of $\text{H"_2}$ using the molar mass of $\text{H"_2}$.

$50.020 \cancel{\text{mol H"_2xx(2.0159"g H"""_2)/(1cancel"mol H"_2)="100.84 g H"""_2}}$