a. For the given combination of reactants, the limiting reagents is F₂
b. The mass of HF that would be made is 106.65 g
Stoichiometry
From the question, we are to identify the limiting reagent and calculate the mass of HF made
First, we will write the balanced chemical equation for the reaction
3F₂ + NH₃ → 3HF + NF₃
This means 3 moles of F₂ reacts with 1 mole of NH₃ to produce 3 moles of HF and 1 mole of NF₃
From the given information,
We have 5.33 moles of F₂ and 2.22 moles of NH₃
Since,
3 moles of F₂ reacts with 1 mole of NH₃
Then,
5.33 moles of F₂ will react with 1.78 moles of NH₃
Therefore, NH₃ is in excess and F₂ is the limiting reagent
Hence, for the given combination of reactants, the limiting reagents is F₂
b. For the mass of HF that would be made
From the balanced chemical equation,
3 moles of F₂ reacts with 1 mole of NH₃ to produce 3 moles of HF
Then,
5.33 moles of F₂ will react with 1.78 moles of NH₃ to produce 5.33 moles of HF
Number of moles of HF produced = 5.33 moles
Using the formula,
Mass = Number of moles × Molar mass
Molar mass of HF = 20.01 g/mol
∴ Mass of HF produced = 5.33 × 20.01
Mass of HF produced = 106.65 g
Hence, the mass of HF that would be made is 106.65 g
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