Question

OWL 2 Assignmen... OWL2 Online tea... [Review Topics) (References) Use the References to access important values if needed for this question. According to the following reaction, how many grams of iron(III) chloride will be formed upon the complete reaction of 30.2 grams of chlorine gas with excess iron? iron(s) + chlorineg) iron(III) chloride(s) grams iron(III) chloride An error has been detected in your answer. Check for typos. miscalculations etc. before submitting your answer Submit Answer Retry Entire Group 3 more group attempts remaining

Answer 1

ANSWER:

Data:

• 30.2 g of Cl₂ (limiting reactant)
• Excess of Fe (excess reactant)
• MW Cl₂ = 70.90 g/mol
• MW FeCl₃ = 162.20 g/mol

The balanced chemical equation is:

$2\, Fe\, (s)+3\, Cl_{2}\, (g)\rightarrow 2\, FeCl_{3}\, (s)$

The amount of FeCl₃ could be calculated from the mass of Cl₂ and several conversion factors using the balanced equation and molar weight values:

$mass\, FeCl_{3}=mass\, Cl_{2} \times \frac{1\, mol\, Cl_{2}}{70.90\, g\, Cl_{2}}\times \frac{2\, mol\, FeCl_{3}}{3\, mol\, Cl_{2}}\times \frac{162.20\, g\, FeCl_{3}}{1\, mol\, FeCl_{3}}$

$mass\, FeCl_{3}=30.2\, g\, Cl_{2} \times \frac{1\, mol\, Cl_{2}}{70.90\, g\, Cl_{2}}\times \frac{2\, mol\, FeCl_{3}}{3\, mol\, Cl_{2}}\times \frac{162.20\, g\, FeCl_{3}}{1\, mol\, FeCl_{3}}$

$mass\, FeCl_{3}=\mathbf{46.06\, g\, FeCl_{3}}$

The amount of FeCl₃ produced from 30.2 g of Cl₂ is 46.06 g