Question

if 38.5 g of no and 26.9 g of o₂ react together, what is the mass, in grams, of no₂ that can be formed via the reaction below? 2no(g) + o₂(g) → 2no₂(g)

Explanation:

Step1: Calculate moles of reactants

• Molar mass of \(NO\): \(M_{NO}=14 + 16=30\space g/mol\). Moles of \(NO\), \(n_{NO}=\frac{38.5\space g}{30\space g/mol}\approx1.283\space mol\)
• Molar mass of \(O_{2}\): \(M_{O_{2}} = 2\times16=32\space g/mol\). Moles of \(O_{2}\), \(n_{O_{2}}=\frac{26.9\space g}{32\space g/mol}\approx0.8406\space mol\)

Step2: Determine limiting reactant

From the balanced equation \(2NO(g)+O_{2}(g)\to2NO_{2}(g)\), the mole ratio of \(NO\) to \(O_{2}\) is \(2:1\).
If \(O_{2}\) is the limiting reactant, moles of \(NO\) required \(n_{NO}^{'}=2\times n_{O_{2}} = 2\times0.8406 = 1.681\space mol\). But we have \(n_{NO}=1.283\space mol\).
If \(NO\) is the limiting reactant, moles of \(O_{2}\) required \(n_{O_{2}}^{''}=\frac{n_{NO}}{2}=\frac{1.283}{2}=0.6415\space mol\). Since \(n_{O_{2}} = 0.8406\space mol>0.6415\space mol\), \(NO\) is the limiting reactant.

Step3: Calculate moles of \(NO_{2}\)

From the balanced equation, mole ratio of \(NO\) to \(NO_{2}\) is \(1:1\). So moles of \(NO_{2}\), \(n_{NO_{2}}=n_{NO}\approx1.283\space mol\)

Step4: Calculate mass of \(NO_{2}\)

Molar mass of \(NO_{2}\): \(M_{NO_{2}}=14+(2\times16)=46\space g/mol\). Mass of \(NO_{2}\), \(m_{NO_{2}}=n_{NO_{2}}\times M_{NO_{2}}=1.283\times46\space g\approx59.0\space g\)

Answer:

\(59.0\space g\)