How many grams of Cl2 can be prepared from the reaction of 1

How many grams of Cl2 can be prepared from the reaction of 16.0 g of MnO2 and 30.0 g of HCl according to the following chemical equation? MnO2+4(HCl) --> MnCl2 + Cl2 + 2(H2O)

Mr of MnO2 = 55+32 = 77 Moles of MnO2 in 16g = mass/mr= 16 / 77 = 0.21 1:1 ratio of MnO2 to Cl2 Moles of Cl2 = 0.21 Mr of Cl2 = 71 Mass = moles x Mr = 0.21 x 71 = 14.8g. Answer = 14.8g