11/13/22

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?
CaF2+H2SO4→ 2HF+CaSO41.5 moles CaF2 will give 3.0 moles HF1 mole HF occupies 22.4 L at STPmoles HF that will occupy 30.2 L at STP= 1.35 moles HF% yield = (1.35 moles/3.0 moles) * 100 = 45%
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