Given the reaction 2 NO + O_2 → 2 NO_2, what mass of O_2 (in grams) is required to produce 75.0 grams of NO_2?

• A. 32.0 g
• B. 48.0 g
• C. 24.6 g
• D. 16.0 g

Answer: (C)

Verified step by step guidance

1. Write down the balanced chemical equation: \$2\ NO + O_2 \rightarrow 2\ NO_2$.
2. Calculate the molar mass of \(NO_2\) by adding the atomic masses of nitrogen and oxygen: \(Molar\ mass\ of\ NO_2 = 14.0\ g/mol + 2 \times 16.0\ g/mol\).
3. Convert the given mass of \(NO_2\) (75.0 g) to moles using the molar mass: \(moles\ of\ NO_2 = \frac{75.0\ g}{Molar\ mass\ of\ NO_2}\).
4. Use the stoichiometric ratio from the balanced equation to find moles of \(O_2\) required. According to the equation, 2 moles of \(NO_2\) are produced per 1 mole of \(O_2\), so: \(moles\ of\ O_2 = \frac{1}{2} \times moles\ of\ NO_2\).
5. Calculate the mass of \(O_2\) needed by multiplying the moles of \(O_2\) by its molar mass (32.0 g/mol): \(mass\ of\ O_2 = moles\ of\ O_2 \times 32.0\ g/mol\).