The image shows a chemistry question that reads, "What mass of O2 is used when 3.75 x 10^24 atoms of Mg are used?"
To answer this question, we need additional information, typically a balanced chemical equation that describes how magnesium (Mg) reacts with oxygen (O2). Assuming that magnesium reacts with oxygen in a simple combustion reaction to form magnesium oxide (MgO), the balanced equation would be:
\[ 2Mg + O_2 → 2MgO \]
This equation tells us that two moles of magnesium react with one mole of oxygen gas to produce two moles of magnesium oxide. From Avogadro's number, we know that 1 mole of any substance contains \( 6.022 \times 10^{23} \) entities (atoms, molecules, etc.). Therefore, we can calculate the moles of magnesium and then use the stoichiometry from the balanced equation to find out the mass of \( O_2 \) that would react with that amount of magnesium.
Moles of Mg = \( \frac{3.75 \times 10^{24} \text{ atoms}}{6.022 \times 10^{23} \text{ atoms/mole}} \)
Moles of Mg ≈ 6.225 moles
According to the balanced equation, 2 moles of Mg react with 1 mole of O2. Therefore:
Moles of \( O_2 \) = \( \frac{6.225 \text{ moles Mg}}{2} \)
Moles of \( O_2 \) ≈ 3.1125 moles
Next, we calculate the mass of \( O_2 \) using its molar mass (32.00 g/mole, since \( O_2 \) has a molecular weight of 16.00 g/mole for each oxygen atom and there are two atoms per molecule):
Mass of \( O_2 \) = moles of \( O_2 \) \(\times\) molar mass of \( O_2 \)
Mass of \( O_2 \) ≈ 3.1125 moles \(\times\) 32.00 g/mole
Mass of \( O_2 \) ≈ 99.60 grams
Therefore, approximately 99.60 grams of \( O_2 \) would be used when 3.75 x 10^24 atoms of Mg are used, assuming the reaction is the simple combustion of magnesium to form magnesium oxide.
