Question

How many moles of NiCl2 can be formed in the reaction of 9.13 mol of Ni(s) and 11.68 mol of HCl(aq)? The other product is H2(g). Report your answer to the hundredths of a mol without units.

Answer 1

9.13 mol of Ni(s) and 11.68 mol of HCl(aq)

Ni(s) + 2HCl(aq) ------------> NiCl2(aq) + H2(g)

1 moles of Ni react with 2 moles of HCl

9.13 mole of Ni react with = 2*9.13/1   = 18.26 moles of HCl is required

HCl is limiting reactant

2 moles of HCl react with excess of Ni to gives 1 mole of NiCl2

11.68 moles of HCl react with excess of Ni to gives = 1*11.68/2   = 5.84moles of NiCl2 >>>>answer

5.84 >>>>answer