Given the reaction 6 I2O5 + 20 BrF3 → 12 IF5 + 15 O2 + 10 Br2, if 44 g of I2O5 are combined with 100 g of BrF3, which is the limiting reactant?

• A. BrF3
• B. I2O5
• C. Neither; both are in excess
• D. Both are limiting reactants

Answer: (B)

Verified step by step guidance

1. Write down the balanced chemical equation: \$6\ I_2O_5 + 20\ BrF_3 \rightarrow 12\ IF_5 + 15\ O_2 + 10\ Br_2$.
2. Calculate the molar masses of the reactants: find the molar mass of \(I_2O_5\) and \(BrF_3\) by summing the atomic masses of their constituent atoms.
3. Convert the given masses of each reactant to moles using the formula: \(\text{moles} = \frac{\text{mass}}{\text{molar mass}}\).
4. Determine the mole ratio of the reactants based on the balanced equation: for every 6 moles of \(I_2O_5\), 20 moles of \(BrF_3\) are required. Calculate the actual mole ratio from the given amounts and compare it to the stoichiometric ratio.
5. Identify the limiting reactant by comparing the mole ratios: the reactant that produces fewer moles of product (or runs out first according to the stoichiometric ratio) is the limiting reactant.