# Question 97035

Oct 8, 2017

64%

#### Explanation:

Start by calculating the theoretical yield of the reaction, i.e. how much hydrogen gas would be produced at 100% yield.

$2 {\text{Na"_ ((s)) + 2"H"_ 2"O"_ ((l)) -> 2"NaOH"_ ((aq)) + "H}}_{2 \left(g\right)} \uparrow$

Notice that for every $2$ moles of sodium that take part in the reaction, the reaction produces $1$ mole of hydrogen gas.

Use the molar mass of sodium metal to convert the mass to moles

0.61 color(red)(cancel(color(black)("g"))) * "1 mole Na"/(22.99color(red)(cancel(color(black)("g")))) = "0.02653 moles Na"

According to the balanced chemical equation, the reaction should produce

0.02653 color(red)(cancel(color(black)("moles Na"))) * "1 mole H"_2/(2color(red)(cancel(color(black)("moles Na")))) = "0.013265 moles H"_2

Use the molar mass of hydrogen gas to convert this to grams

0.013265 color(red)(cancel(color(black)("moles H"_2))) * "2.016 g"/(1color(red)(cancel(color(black)("mole H"_2)))) = "0.02674 g"

So, you know that at 100% yield, the reaction produces $\text{0.02674 g}$ of hydrogen gas when $\text{0.61 g}$ of sodium metal take part in the reaction.

However, you know that the actual yield of the reaction is $\text{0.017 g}$ of hydrogen gas.

To find the percent yield of the reaction, which tells you how many grams of product are actually produced for every $\text{100 g}$ of product that could theoretically be produced, use the equation

"% yield" = "actual yield"/"theoretical yield" * 100%

In your case, you will have

"% yield" = (0.017 color(red)(cancel(color(black)("g"))))/(0.02674color(red)(cancel(color(black)("g")))) * 100% = color(darkgreen)(ul(color(black)(64%)))#

The answer is rounded to two sig figs.