Question

For each of the following unbalanced reactions, sup- pose exactly 5.00 g of each reactant is taken. Deter- mine which reactant is limiting, and also determine what mass of the excess reagent will remain after the limiting reactant is consumed.

Answer 1

Answer:

The reactant that is limiting is the Na2B4O7, and the mass of the excess reagent (H2SO4) is 2.5627g.

Explanation:

The unbalanced reaction is:

Na2B4O7 + H2SO4 + H2O ⇒ H3BO3 + Na2SO4

• First the equation is balanced

Na2B4O7 + H2SO4 + 5H2O ⇒ 4H3BO3 + Na2SO4

According to the balanced equation we observe that it takes 1 mol of Na2B4O7 to react with 1 mol of H2SO4.

• Then the molecular weights of each reactant are determined:

MWNa2B4O7 = 201.197 g/mol

MWH2SO4 = 98.075 g/mol

• It is calculated how many moles will react with 5 initial grams of each.

201.197 g of Na2B4O7 ⇒ 1 mol

5 g                                 ⇒ X

$X=\frac{5g*1mol}{201.197g/mol} =0.02485mol$ of Na2B4O7

98.075 g of H2SO4 ⇒ 1 mol

5 g                             ⇒ X

$X=\frac{5g*1mol}{98.075g/mol} =0.05098mol$ of H2SO4

The limiting reagent is the one that is consumed first, in this case it is Na2B4O7, since 0.02485 moles are needed for both reactants, then there are 0.02613 moles of H2SO4 left over. With these moles the grams of excess reagent are calculated.

$mass=MW*mol=98.075g/mol*0.02613mol=2.5627g$