Final answer:
To determine the limiting reactant between Sn and HF, we convert the mass of each reactant to moles and compare the mole ratio from the balanced equation. Sn would require more HF than we have, making HF the limiting reactant.
Step-by-step explanation:
To determine which reactant is limiting in the reaction Sn (s) + 2HF (g) → SnF₂ (s) + H₂(g), we first need to calculate how many moles of each reactant we have. The molar mass of Sn is approximately 118.71 g/mol.
For Sn, we have:
91.7 g Sn * (1 mol Sn / 118.71 g Sn) = 0.773 mol Sn
The balanced equation shows that 1 mole of Sn reacts with 2 moles of HF. So, 0.773 moles of Sn would require 1.546 moles of HF to react completely:
0.773 mol Sn * (2 mol HF / 1 mol Sn) = 1.546 mol HF
To find out how many moles of HF we have, we also use its molar mass which is 20.01 g/mol.
For HF, we have:
15.8 g HF * (1 mol HF / 20.01 g HF) = 0.790 mol HF
Since we need 1.546 moles of HF but only have 0.790 moles, HF is the limiting reactant.