Answer to Question #267009 in Inorganic Chemistry for rei

Question

If 0.05 mol of Zinc was added to 0.075 mol of HCl in the given equation, Zn(s) + 2HCl(aq)  ZnCl2(aq) + H2(g) (b) How many moles of the excess reactant did not react/ excess? (3pts) *

Answer 1

Zn:ZnCl2= 1:1

=0.05mol

HCl: ZnCl2 =2:1

= 0.075/2= 0.0375mole

= 0.0875mole of ZnCl2