Question

A 0.271 g piece of solid magnesium reacts with gaseous oxygen from the atmosphere to form solid magnesium oxide. In the laboratory a student weighs the mass of the magnesium oxide collected from this reaction as 0.234 g.

What is the percent yield of this reaction?

Answer 1

1)

Molar mass of Mg = 24.31 g/mol

mass(Mg)= 0.271 g

use:

number of mol of Mg,

n = mass of Mg/molar mass of Mg

=(0.271 g)/(24.31 g/mol)

= 1.115*10^-2 mol

Balanced chemical equation is:

2 Mg + O2 ---> 2 MgO

Molar mass of MgO,

MM = 1*MM(Mg) + 1*MM(O)

= 1*24.31 + 1*16.0

= 40.31 g/mol

According to balanced equation

mol of MgO formed = (2/2)* moles of Mg

= (2/2)*0.0111

= 0.0111 mol

use:

mass of MgO = number of mol * molar mass

= 0.0111*40.31

= 0.4494 g

% yield = actual mass*100/theoretical mass

= 0.234*100/0.4494

= 52.0737 %

Answer: 52.1 %