I think you must mean 7.42 x 10^24 molecules of HF, since I doubt that there are that many grams available.
since each atom of Sn reacts with 2 molecules of HF, it would require half as many Sn as HF.
How many atoms of tin are necessary to
react completely with 7.42 x 10(24th)g
of HF.
Sn(s) + 2HF(g) = SnF2(S) + H2 (g)
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