Question

A reaction vessel contains 8.00g NaCl and 8.00g H2SO4.

a.) What is the limiting reactant

b.) How many grams of HCl was produced

c.) How many grams of excess reactant are left unreacted/unconsumed

Answer 1

Sol:(a)moles of NaCl=mass/molar mass

=8.00g/58.44g/mol=0.137mol

Moles of H2SO4=8.00g/98g/mol=0.082mol

Reaction between NaCl and H2SO4 is:

2NaCl+H2SO4 -> 2HCl +Na2SO4

From equation, one mole of H2SO4 react with 2mol of NaCl to give product.

Hence 0.137mol NaCl produce 0.137mol HCl

And 0.082mol H2SO4 produce 0.164mol HCl

Hence NaCl is limiting reagrnt since it give smaller number of moles of HCl.

(b) moles of NaCl=moles of HCl produced

Hence moles of HCl produced =0.137mol

Mass of HCl produced=0.137mol×36.5g/mol=5.00g HCl

(c) H2SO4 is the reagent which is in excess.

According to equation :

Moles of H2SO4=1/2(Moles of NaCl)

Hence moles of H2SO4=1/2(0.137) =0.0685mol

Mass of H2SO4=0.0685mol×98g/mol=6.713g

Hence grams of H2SO4 left unreacted=8.00g-6.713g=1.287g