Question

How many grams of Cl2 can be prepared from the reaction of 16.0g of MnO_(2) and 30.0g HCl according to the following chemical equation? MnO_(2) + 4 HCI -> MnCI_(2 )+CI_(2) +2H_(2)O

Answer 1

$\nu(HCl)=30g/(35.5+1)g/mol=0.822mol$

$\nu(MnO_2)=16g/(55+2*16)g/mol=0.184mol$

That is, acid in excess. Then, following the equation $\nu(MnO_2)=\nu(Cl_2)$

$=>m(Cl_2)=\nu(Cl_2)M(Cl_2)=0.184*71g=13.064g$