Question

During an investigation, a student burns magnesium to form magnesium oxide. the starting mass is measured at 21.3g if the student recovers 30.2g of magnesium oxide what is the percent yield?

A)63.4%
B)70.5%
C)74.9%
D)86.0%

Answer 1

The percent yield is `"85.6%"`.

I would go with `"D"`.

Explanation:

Balanced equation

`"2Mg(s) + O"_2("g")"` `rarr` `"2MgO"`

We will use stoichiometry to determine the theoretical yield of magnesium oxide. Then we will use the percent yield equation to determine the percent yield of magnesium oxide.

The process will go as follows:

`color(red)"mass Mg"` `rarr` `color(green)"mol Mg"` `rarr` `color(blue)"mol MgO"` `rarr` `color(purple)"mass MgO"`

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

`color(red)"mass Mg"` `rarr` `color(green)"mol Mg"`

Divide the given mass of magnesium by its molar mass `("24.305 g/mol")`. Do this by multiplying by the reciprocal of the molar mass, mol/g.

`21.3color(red)cancel(color(black)("g Mg"))xx(1"mol Mg")/(24.305color(red)cancel(color(black)("g Mg")))="0.876 mol Mg"`

`color(green)"mol Mg"` `rarr` `color(blue)"mol MgO"`

Multiply mol `"Mg"` by the mol ratio between `"MgO"` and `"Mg"` from the balanced equation, with `"MgO"` in the numerator.

`0.876color(red)cancel(color(black)("mol Mg"))xx(2"mol MgO")/(2color(red)cancel(color(black)("mol Mg")))="0.876 MgO"`

`color(blue)"mol MgO"` `rarr` `color(purple)"mass MgO"`

Multiply mol `"MgO"` by its molar mass `("40.304 g/mol")`.

`0.876color(red)cancel(color(black)("mol MgO"))xx(40.304"g MgO")/(1color(red)cancel(color(black)("mol MgO")))="35.3 g MgO"`

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

`"% yield"=("actual yield")/("theoretical yield")xx100%`

`"actual yield"` `=` `"30.2 g"`

`"theoretical yield"` `=` `"35.3 g"`

`"% yield"=(30.2color(red)cancel(color(black)("g")))/(35.3color(red)cancel(color(black)("g")))xx100="85.6 %"`

The percent yield is `"85.6%"`.