Larutan NaOH 0,1 M sebanyak 200 mL direaksikan dengan 200 mL lauratan MgCI 2 0,1 M sehingga terjadi reaksi : NaOH( aq) + MgCI 2 ( aq ) --> Mg(OH) 2 ( s ) + NaCL( aq ) Jika dikerahui A r Mg = 24, O = 16 dan H = 1, jumlah maksimum endapan yang dihasilkan adalah.... A. 58,00 B. 11,60 C. 5,80 D. 1,16 E. 0,58

<p><strong>Pembahasan</strong></p><p>Diketahui:&nbsp;</p><p>200 mL NaOH 0.1M = 20 mmol</p><p>200 mL MgCl⁠⁠⁠⁠⁠⁠⁠₂0.1M = 20 mmol</p><p>&nbsp;</p><p>Reaksi yg terjadi (sdh disetarakan):</p><p>2 NaOH(aq) + MgCI₂(aq) --&gt; Mg(OH)₂(s) + NaCl(aq)&nbsp;</p><p>&nbsp; 20 mmol &nbsp; &nbsp; &nbsp; 20 mmol &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp;(awal)&nbsp;</p><p>&nbsp; 20 mmol &nbsp; &nbsp; &nbsp; 10 mmol &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; 10 mmol &nbsp; &nbsp; &nbsp;(<i>reaksi)&nbsp;</i></p><p>&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp;</p><p>Perhatikan reaksi setara tsb, ternyata sbg pembatas reaksi adalah NaOH sebesar 20 mmol sehingga endapan Mg(OH)₂ yg terbentuk sebesar 10 mmol</p><p>Endapan yg dihasilkan: 10/1000 x 58 = 0.58 g</p><p>&nbsp;</p><p>Jawaban &nbsp;<strong>E</strong></p><p>&nbsp;</p>