Answer to Question #121620 in General Chemistry for queen

Question

During the reaction of 250 mL of 4.50 mol/L of hydrochloric acid with 5.13g of sulfur, ____i____ will be the limiting reagent and ____ii____ will be the number of moles of chlorine gas produced


a) i= hydrochloric acid; ii= 1.125 mol

b) i= hydrochloric acid; ii= 0.5625 mol

c) i= sulfur ; ii=0.01999mol

d) i=sulfur; ii=0.1599 mol

Answer 1

$2HCl+ S \longrightarrow H_2S+ Cl_2$

For HCl,

$Molarity= \frac{moles}{volume(in L)}$

$moles=4.5\times .25= 1.125,$

mass of HCl=$1.125\times 36.5=41.0625 g$

2 mole of HCl react with 1 mole HCl , and here moles of S=$\frac{5.13}{32}=0.16 mole$

so here limiting reagent is S and

amount of HCl needed= .32 moles of HCl

and ,

mass of HCl = 11.68 g

and moles of Cl₂ is = 0.16 moles