Hidrogen Fluorida (HF) merupakan gas yang dimanfaatkan untuk membuat freon. Gas HF dapat dihasilkan melalui reaksi sebagai berikut: CaF 2 (s) + H 2 SO 4 (aq) --> CaSO 4 (s) + 2HF (g) Jika rendemen reaksinya adalah 90% maka untuk mendapatkan 22,4 liter gas HF (0°, 1 atm), massa CaF 2 yang harus direaksikan adalah..... (Ar) Ca = 40 dan F = 19

<p>Untuk menjawab soal ini dapat dilakukan dengan menggukan konsep stoikiometri Mula Reaksi dan Setimbang.</p><p>&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp;&nbsp;<strong>CaF_2^(s)+ H₂SO_4^(aq) --&gt; CaSO_4^(s) + 2HF_^(g)</strong></p><p>Mula &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp;0,55 mol</p><p>Reaksi &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; 0,5 mol &nbsp; 0,5 mol &nbsp; &nbsp; &nbsp;0,5 mol &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; 1 mol</p><p>Setimbang &nbsp; &nbsp; &nbsp; 0,05mol &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp;1 mol</p><p>&nbsp;</p><p>Gas HF 22,4L (0^oC, 1 atm) keadaan STP Rendemen 90%</p><p>Mr HF = 1 + 19 = 20 gr/mol</p><p>Mr CaF₂ = 40 + (19x2) = 78 gr/mol</p><p>&nbsp;</p><p>mol HF = 22,4L/22,4L/mol(STP) = 1 mol (dimasukkan ke persamaan MRS)</p><p>maka mula-mula ada <strong>0,55 mol CaF₂.</strong></p><p><strong>massa CaF₂ = 0,55 mol x 78 gr/mol = 42,9 gram</strong></p><p>&nbsp;</p>