If 0.86 mole of MnO2 and 48.2 g of HCL react, which reagent will be used up first? How many grams of Cl2 will be produced? Calculate the number of grams of the excess reagent remaining at the end of the reaction

MnO2 + 4HCl = MnCl2 + Cl2 + 2H2O

m (HCl) = 36.45/48.2 = 0.76

every 4 moles of HCl will only react with 1 mole MnO2 and there is less HCl therefore HCl will be used up first.

Excess MnO2 = 0.86 - (0.76/4) = 0.67m

grams MnO2 = 0.67x 86.93 = 58.24grams left over.

Every 4 moles HCl will only form 1 mole Cl2 so m (Cl2) = 0.76/4=0.19

0.19x70.9=13.47grams Chlorine formed