Question #97035
1 Answer
Explanation:
Start by calculating the theoretical yield of the reaction, i.e. how much hydrogen gas would be produced at
#2"Na"_ ((s)) + 2"H"_ 2"O"_ ((l)) -> 2"NaOH"_ ((aq)) + "H"_ (2(g)) uarr#
Notice that for every
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
However, you know that the actual yield of the reaction is
To find the percent yield of the reaction, which tells you how many grams of product are actually produced for every
#"% 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.