# Question 748a3

Nov 25, 2016

Theoretical yield = 0.10 g of hydrogen gas. % yield = 80%

#### Explanation:

The first thing you need to do when you see this sort of problem is write down the balanced chemical equation. In this case, you are reacting water with sodium metal. In general, alkali metals will react with water to produce the alkali metal hydroxide (and hydrogen):

$2 N {a}_{\text{(s)" + 2 H_2O_"(l)" rarr 2NaOH_"(aq)" + H_2}} \left(g\right)$

From this you get the mole ratio of sodium to hydrogen gas by examining the coefficients of the balanced chemical equation:

2 moles of sodium metal will yield one mole of hydrogen gas.

$2.3 \text{g " Na xx (1 "mole " Na)/(22.989769 "g " Na) = 0.1000445 "moles } N a$

We therefore will produce:

 0.1000445 "moles " Na xx (1 "mole " H_(2"(g)"))/(2 "moles " Na) = 0.050022 "mole "H_(2"(g)")

And then, finally, How many grams of hydrogen is that?

0.050022 "mole "H_(2"(g)") xx (2.016 "g " H_(2"(g)"))/(1 "mole "H_(2"(g)")) = 0.100844 "g "H_(2"(g)")

Now we can only use 2 significant figures, so the final theoretical yield is 0.10 g of hydrogen gas.

The % yield is simply the actual yield (0.080 g) divided by the theoretical yield (0.10 g) multiplied by 100 %:

(0.08 "g "H_(2"(g)"))/(0.10 "g "H_(2"(g)")) xx 100% = 80%

Nov 25, 2016

Theoretical Yield is 0.10g
The Percent Yield is 79%

#### Explanation:

First, We know that

$2 N {a}_{\left(s\right)} + {H}_{2} O \to {H}_{2} + N {a}_{2} O$

${M}_{N a} = 22.9898 \frac{g}{m o l}$

${M}_{{H}_{2}} = 2.0158 \frac{g}{m o l}$

So the Sodium to Hydrogen gas $\left({H}_{2}\right)$ Mole ratio (derived from the balanced chemical equation is 2 to 1 respectively.

First lets convert our Sodium to Moles.

(2.3cancel"g")/(22.9898 cancel"g"/(mol))=0.1000443675 "mol"

Now we will apply the Mole ratio to figure out the theoretical yield (in mol) of the Hydrogen gas.

$0.1000443675 \cancel{\text{mol of Na"*(1"mol of "H_(2)"")/(2cancel"mol of Na") = 0.05002218375 "mol of"H_(2)}}$

Lets turn this into grams

$0.05002218375 \cancel{\text{mol"*2.0158"g"/cancel"mol"=0.100834718"g}}$

So 0.10g is the Theoretical yield (with significant figures [2] )

The percent yield is the actual (experimental) yield divided by the theoretical.

(0.080"g")/(0.100834718"g")=0.79337753488 *100%=79.3%#

The percent yield is 79% with 2 sig-figs. Your teacher may say it is 80%; I would tell him/her to not apply sig-figs until the answer.