Question

1.0 g of Mg is burnt with 0.28 g of ${{O}_{2}}$ in a closed vessel. Which reactant is left in excess and how much?
A. Mg, 5.8 g
B. Mg, 0.58 g
C. ${{O}_{2}}$, 0.24g
D. ${{O}_{2}}$, 2.4 g

Answer 1

Hint: The reactant that fully reacts in the reaction is called reactant limiting or reagent limiting. The reactant that is not fully consumed in the reaction is called excess reactant. You should write the balanced reaction equation before solving the question. A balanced equation is an equation for a chemical reaction in which the number of atoms in the reaction for each element and the total charge for both the reactants and the products is equal.

Complete answer:
The Law of Definite Proportions, states that any chemical compound will always contain a fixed ratio of elements by mass. The Law of Definite Proportions is also sometimes called Proust's Law.
Mg is burnt with ${{O}_{2}}$ in a closed vessel to produce MgO.

The reaction used in question is: $2Mg+{{O}_{2}}\to 2MgO$
1 mole of ${{O}_{2}}$ reacts with 2 moles of Mg to produce 2 moles of MgO.
Moles (n) = $\dfrac{given weight(w)}{molar mass(m)}$
As given in question, the weight of ${{O}_{2}}$ is 0.28g.
Molar mass of ${{O}_{2}}$ is 32.
Therefore, numbers of moles of ${{O}_{2}}$ $\dfrac{0.28}{32}=0.00875$
0.00875 moles of ${{O}_{2}}$ reacts with $2\times 0.00875$ moles of Mg $=0.0175$ moles of Mg to produce $=0.0175$ moles of MgO.
Mass of magnesium that reacts = $moles\times molar mass$
Molar mass of Mg is 24.
Hence, mass of magnesium that reacts $0.0175\times 24=0.42$
That means, out of the 1g of Mg, only 0.42g is used.
Therefore, Mg is in excess by (1 - 0.42) = 0.58 g.
Therefore, limiting reagent is ${{O}_{2}}$.
So, the correct answer is “Option B”.

Note: The limiting reagent (also known as limiting reactant) in a chemical reaction is a reactant that is totally consumed when the chemical reaction is completed. The amount of product formed is limited by this reagent, since the reaction cannot continue without it.
Remember this reaction equation,
$2Mg+{{O}_{2}}\to 2MgO$
When 2 moles of Mg combine with 1 mole of ${{O}_{2}}$ to produce 2 moles of MgO.