Question

how many grams of molecular chlorine will be required to completely react with 0.0223 mol of sodium iodide according to the following reaction? 2nai + cl₂ → 2nacl + i₂

Explanation:

Step1: Determine the mole ratio

From the balanced equation \(2NaI + Cl_{2}\to2NaCl + I_{2}\), the mole ratio of \(NaI\) to \(Cl_{2}\) is \(2:1\).

Step2: Calculate moles of \(Cl_{2}\)

Given \(n(NaI)=0.0223\ mol\). Using the mole ratio \(\frac{n(Cl_{2})}{n(NaI)}=\frac{1}{2}\), so \(n(Cl_{2})=\frac{0.0223\ mol}{2}= 0.01115\ mol\).

Step3: Calculate mass of \(Cl_{2}\)

The molar mass of \(Cl_{2}\) is \(M = 2\times35.45\ g/mol=70.9\ g/mol\). Using the formula \(m = n\times M\), \(m(Cl_{2})=0.01115\ mol\times70.9\ g/mol\approx0.791\ g\).

Answer:

\(0.791\ g\)