Question

Magnesium oxide can be made by heating magnesium metal in the presence of oxygen. The balanced equation for the reaction is:
2Mg(s)+O2(g)-->2MgO(s)
When 10.1g of Mg are allowed to react with 10.5g of O2 11.0g of MgO are collected.

A. determine the limiting reactant for the reaction.

B. determine the theoretical yeild for the reaction

C. determine the percent yeild for the reaction.

B.

Answer 1

2 Mg(s) + O₂(g) 2 MgO(s)

Molar mass of Mg = 24.305 g/mol

So, 24.305 g of Mg = 1 mol

10.1 g of Mg = (10.1 / 24.305) mol = 0.416 mol

Molar mass of O₂ =16 g/mol

So, 16 g of O₂ = 1 mol

10.5 g of O₂ = (10.5 / 16) mol = 0.656 mol

Molar mass of MgO = 40.304 g/mol

So, 40.304 g of MgO = 1 mol

11 g of MgO = (11 / 40.304) mol = 0.273 mol.

In this chemical reaction;

2 moles of Mg reacts with 1 mole of O₂ produces 2 moles of MgO.

1 mole of Mg reacts with 1/2 mole of O₂ produces 1 moles of MgO.

A. O₂ is the limiting reactant for the reaction.

B. 0.656 mol of O₂ will produce 1.312 mol of MgO.

Now,

1 mole of MgO = 40.304 g

1.312 mol of MgO = (1.312 x 40.304) g = 52.88 g

So, theoritical yield = 52.88 g

C.

Theoritical yield = 52.88 g

Experimental yield = 11.0 g

So,

% Yield = [(Experimental yield) / (Theoritical yield) ] x 100

= (11 / 52.88) x 100

= 20.8 %