Question #97035

1 Answer
Oct 8, 2017

Answer:

#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.