A 0.301 g piece of solid Mn reacts with gaseous oxygen to form solid magnesium oxide.a student weighs the mass of the magn. oxide collected from this reaction as 0.239 g. what is the percent yield

Answer 1

First, write the balanced equation for the reaction:

Mn(s) + O2(g) ==> MnO(s)

The O2 will be in excess as the problem does not state a specific amount of O2.

Find the moles of Mn present:

0.301 g Mn x 1 mol Mn / 54.94 g = 5.48x10^-3 moles Mn

Next, determine the theoretical yield of MnO:

5.48x10^-3 moles Mn x 1 mol MnO / 1 mol Mn = 5.48x10^-3 moles MnO x 70.9 g MnO/mol = 0.388 g MnO

Finally, calculate % yield;

% yield = actual yield / theoretical yield (x100%) = 0.239 g / 0.388 g (x100%) = 61.6% yield