Question

1.0 g of magnesium is burnt with 0.56 g O₂ in a closed vessel. Which reactant is left in excess and how much? (At. wt. Mg = 24 ; O = 16)

• A. Mg, 0.16 g
• B. O_2, 0.16 g
• C. Mg, 0.44 g
• D. O₂, 0.28 g

Hint:

The equation is \(2Mg + O_2 → 2MgO\)

Answer 1

The Correct Option is A

The balanced chemical equation is \(\underset{\begin{smallmatrix} \\ \\ 24g \end{smallmatrix}}{\mathop{Mg}}\,+\underset{\begin{smallmatrix} \\ 16g \end{smallmatrix}}{\mathop{\frac{1}{2}{{O}_{2}}}}\,\xrightarrow{\,}\underset{\begin{smallmatrix} \\ \\ 40g \end{smallmatrix}}{\mathop{MgO}}\,\) 
From the above equation, we note that 24 g Mg reacts with 16 g O₂ 
Therefore, 1.0 g Mg reacts with \(\frac{16}{24}\times 0.67g\,{{O}_{2}}=0.67g\,{{O}_{2}}.\) 
However, only 0.56 g of O₂ is available.
Thus, O₂ is the limiting reagent.
16 g O₂ then reacts with 24 g Mg. 
\(\therefore\) 0.56 g O₂ will react with Mg \(=\frac{24}{16}\times 0.56=0.84g\)
The amount of Mg left unreacted = 1.0 - 0.84g Mg = 0.16g Mg.
Hence, Mg is present in excess and 0.16 g Mg is left behind unreacted.

So, the correct option is (A): Mg, 0.16 g

Answer 2

The equation is \(2Mg + O_2 → 2MgO\)
                            (s)      (g)        (s)
Hence, 2 moles of Magnesium react with 1 mole of \(O_2\) to form 2 moles of MgO.

No. of moles = \(\frac {\text {Mass}}{\text {Molar\ Mass}}\)

• \(Mg \)= \(\frac {1}{24}\) = 0.041667
• \(O_2\) = \(\frac {0.56}{32}\) = 0.0175

The mole ratio of Mg: \(O_2\) = 2:1
This means that 2 moles of Mg react with 1 mole of \(O_2\)
Mass = Moles x Molar mass
= 24 x 0.035
= 0.84 g
Therefore, out of 1 gram of magnesium that reacts only 0.84 grams is used.
Hence, when 1.0 g of magnesium is burnt with 0.56 g \(O_2\) in a closed vessel, magnesium is present in excess by 0.16.

So, the correct option is (A): Mg, 0.16 g