Question

Question #97035

Answer 1

`64%`

Explanation:

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

`2"Na"_ ((s)) + 2"H"_ 2"O"_ ((l)) -> 2"NaOH"_ ((aq)) + "H"_ (2(g)) uarr`

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 `"0.02674 g"` of hydrogen gas when `"0.61 g"` of sodium metal take part in the reaction.

However, you know that the actual yield of the reaction is `"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 `"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.