Question

What would the limiting reagent be for the reacton between 1.33g of sodium bromide, 0.8g of i-butanol, and 2g of concentrated H2SO4?

Answer 1

Given :

NaBr = 1.33 g

I butanol = 0.8 g

H2SO4 = 2 g

When we allow above reactant then we get bromobutane from butanol.

Lets show its chemical reaction

2NaBr +2 ibutanol + H2SO4 -- > Na2SO4 +2 bromobutane + 2 H2O

Mol ratio of NaBr to butanol to H2SO4 is 2 : 2 : 1

Calculation of moles of NaBr

Mol = mass in g /molar mass

n NaBr = 1.33 g x 1 mol / 102.894 g

=0.01293 mol

n butanol = 0.8 g x 1 mol / 74.12 g

= 0.0179 mol

n H2SO4 = 2 g x 1 mol / 98.079 g

= .0204 mol H2SO4

Lets find moles of any product from all of these reactant

Lets calculate moles of bromobutane

From NaBr

n bromobutane = n NaBr x 2 mol Bromobutane / 2 mol NaBr

=0.01293 mol n bromobutane

From H2SO4

= 0.0204 mol H2SO4 x 2 mol bromobutane / 1 mol H2SO4

=0.0407 mol bromobutane

From butanol

= 0.0179 mol x 2 mol bromobutane /2 mol butanol

= 0.0179 mol Bromobutane

Moles of Bromobutane from NaBr are less so NaBr is limiting reactant since product formation depends totally on limiting reactant.