Question

Question 2 4 pts Always show your work. No credit will be given for any answers (even if they are correct) for which work is not shown. How many grams of Cly can be prepared from the reaction of 16.0 g of MnO, and 30.0 g of HCl according to the following chemical equation? MnO2 + 4HCI ---> MnCl2 + Cl2 + 2H20 HTML Editora B I VA - A - IKE 23 11xx, E - O VX Gr T To 12pt - Pé O word

Answer 1

Question 2 : The e quation, given balanced chemical Ano it u HCl record Moel, tel2 + 2H₂0 169 300g Malarmass 86.91 glmal 26.46 mol 70.906 g/mol To find limiting reactant: groles of und groles of tee = 300 g 86.11g mee so 184 mee According to 361469m stoichiometric ration 0.8228 male 0. 18u mol gano x I reaction 1 mole = 0.184 -0, 8228 mol ut ce x moz 1 reaction a mol Hee d. 2057 0.184 < 0.2057 Mno, He anon will be the a will be the and limitin Toalhan eness neae tann manaz will be completely reaction mixture the unsumed

And ther approach - To grams at calutate miskrenom moles at of one reactant (say A needed to completely a second realtant (say B). It the stoichio -metric available amount of reactant A exceeds What is needed, then it is excess reactant. If not, then the reactant Br Winke the limiting reactant. Starting with grams of Hcl 16.0 g Moz. Mno,, we calcolate minim needed to completely consumed 16.0g norX Imole MnO 86191 q Mnog 4 moleth 36.469.hth i mole na I molette X = 36.84 g och But, Boog available to us. is monet e 26.846 g Ace needed. a ice be the excess realtana and rano will be limiting reactant. Ocrams of cle can be prepared 16.09 Mnox Imole sama 8 6.918 gana 1 mole cl 70.906 gek Imole no Imole cha (13.052