Given the reaction 2 NO(g) → N2(g) + O2(g), how many grams of NO are required to produce 145 g of N2?

• A. 198 g
• B. 290 g
• C. 145 g
• D. 87 g

Answer: (A)

Verified step by step guidance

1. Write down the balanced chemical equation: \$2\ NO(g) \rightarrow N_2(g) + O_2(g)$.
2. Determine the molar mass of \(N_2\) by adding the atomic masses of two nitrogen atoms: \(M_{N_2} = 2 \times 14.01\ \text{g/mol}\).
3. Calculate the number of moles of \(N_2\) produced using the given mass: \(\text{moles of } N_2 = \frac{145\ \text{g}}{M_{N_2}}\).
4. Use the stoichiometric ratio from the balanced equation to find the moles of \(NO\) required. According to the equation, 2 moles of \(NO\) produce 1 mole of \(N_2\), so \(\text{moles of } NO = 2 \times \text{moles of } N_2\).
5. Calculate the mass of \(NO\) needed by multiplying the moles of \(NO\) by its molar mass (\(M_{NO} = 14.01 + 16.00\ \text{g/mol}\)): \(\text{mass of } NO = \text{moles of } NO \times M_{NO}\).