when 117.2g CaF2 is reacted with excess H2SO4. 30.2 L of HF is recovered at STP. what is the % yield of this reaction process?

Question

CaF2+H2SO4→ 2HF+CaSO4

1.5 moles CaF2 will give 3.0 moles HF

1 mole HF occupies 22.4 L at STP

moles HF that will occupy 30.2 L at STP= 1.35 moles HF

% yield = (1.35 moles/3.0 moles) * 100 = 45%

Answer 1

Spot on approach and correct steps. Just to add a bit more context:

1. The balanced reaction tells us 1 mole of CaF₂ makes 2 moles of HF, so with 1.5 moles of CaF₂ (117.2 g ÷ 78.07 g/mol), you’d expect 3.0 moles of HF if everything went perfectly (theoretical value).
2. At STP, 1 mole of any gas takes up 22.4 L, so 30.2 L of HF means you actually got about 1.35 moles (actual value).
3. Then you just do actual ÷ theoretical: 1.35 ÷ 3.0 = 0.45 → 45% yield.

You clearly understood the stoichiometry, applied gas laws correctly, and nailed the percent yield. Nice work :)