Reference no: EM13332156

MnO2(s) + 4HCl(aq) ? MnCl2(aq) + Cl2(g) + 2H2O(l) In the above reaction 2.43 g of MnO2 and 5.80 g of HCl react to form chlorine gas. ok so i know that i asked this question before but i still keep only getting people telling me to stop asking questions but i still do not understand how to do this question a. How many moles of Cl2 can be produced from the given mass of MnO2? b. How many moles of Cl2 can be produced from the given mass of HCl? c. Which reactant is the limiting reactant? d. What is the maximum amount of Cl2 that can be produced by this reaction? e. Calculate the amount of excess reactant. ok so i got this answer for the question but i still do not understand how to use this The mole ratio of MnO2 to Cl2 is 1:1 so first we calculate the number of moles of MnO2 = 12/(55 + 16 + 16) = 12/87 = 0.1379 moles The same number of moles of Cl2 will be formed. At S.T.P, I mole of Cl2 will occupy 22.4 L so 0.1379 moles will occupy = 3.0897 L but this is at S.T.P so we convert it to the given conditions using PV/T = constant 1 x 3.0897/273 = 0.95 x V/298 V = 3.55 L ok i finally get the start of part A. and B both both = 0.1379 moles but after that i got confused on separating the ques?---> At S.T.P, I mole of Cl2 will occupy 22.4 L so 0.1379 moles will occupy = 3.0897 L but this is at S.T.P so we convert it to the given conditions using PV/T = constant 1 x 3.0897/273 = 0.95 x V/298 V = 3.55 L what is that used for ? can you separate them by the letters ...and then explain it to me by parts because i am still really confused on part C D and E Please do not answer if you are just going to re-answer the question with the same retarded answer ...because i am trying to learn how to do this and i have been trying to work it out all night. And my teacher made this an open class question not a homework question but i need to learn this for a test.