Question

nitric oxide (no) reacts with oxygen gas to form nitrogen dioxide (no₂), a dark brown gas:
2no(g) + o₂(g) → 2no₂(g)
in one experiment, 0.877 mol of no is mixed with 0.509 mol of o₂. determine which of the two reactants is the limiting reactant. calculate also the number of moles of no₂ produced.
round your answer to 3 significant digits.
limiting reactant
moles of no₂ produced

Explanation:

Step1: Calculate moles of \(O_2\) needed if \(NO\) is limiting

From the balanced equation \(2NO(g)+O_2(g)\to 2NO_2(g)\), the mole ratio of \(NO\) to \(O_2\) is \(2:1\).
If \(n(NO) = 0.877\space mol\), then moles of \(O_2\) needed \(n_{required}(O_2)=\frac{0.877\space mol}{2}=0.4385\space mol\)

Step2: Compare required and available \(O_2\)

Available \(n(O_2) = 0.509\space mol\). Since \(0.4385\space mol<0.509\space mol\), \(NO\) is the limiting reactant.

Step3: Calculate moles of \(NO_2\) produced

From the mole ratio of \(NO\) to \(NO_2\) (\(2:2 = 1:1\)), if \(n(NO)=0.877\space mol\), then \(n(NO_2)=0.877\space mol\)

Answer:

Limiting reactant: \(NO\)
Moles of \(NO_2\) produced: \(0.877\) moles