Question

1.5 mol of $O_2$ combine with Mg to form oxide MgO. The mass of Mg (at. mass 24) that has combined is

• A. 72 g
• B. 36 g
• C. 48 g
• D. 24 g

Answer 1

The Correct Option is A

$\underset{\text{48 g}}{ {2 Mg}}+\underset{\text{1 mol}}{ {O_{2}}}\to 2MgO$ Thus, $1.5$ mol of $O_{2}$ combine with $Mg$ $=48 \times 1.5\,g $ $=72\,g$

Answer 2

Given:
Moles of \(O_2\) = 1.5 mol
\(2 \text{ Mg} + O_2 \rightarrow 2 \text{ MgO}\)

Determine the stoichiometry from the equation:
2 moles of Mg react with 1 mole of \(O_2\) to produce 2 moles of MgO.

Calculate moles of Mg required using the stoichiometry:
\(\text{Moles of Mg} = \frac{1.5 \text{ mol } O_2}{1 \text{ mol } O_2} \times 2 \text{ mol } Mg\)
\(\text{Moles of Mg} = 3 \text{ mol } Mg\)

Calculate the mass of Mg:
Atomic mass of Mg = 24 g/mol
\(\text{Mass of Mg} = 3 \text{ mol } Mg \times 24 \text{ g/mol}\)
\(\text{Mass of Mg} = 72 \text{ g}\)

So, the correct option is (A): 72g