Question

For each of the following unbalanced reactions, suppose 5.18 g of each reactant is taken. Determine which reactant is limiting, and also determine what mass of the excess reagent(s) will remain after the limiting reactant is consumed. Na2B4O7(s) + H2SO4(aq) + H2O(l) → H3BO3(s) + Na2SO4(aq)

Answer 1

first balance the equation

Na₂B₄O₇(s) + H₂SO₄(aq) +5H₂O(l) ------> 4H₃BO₃(s) + Na₂SO₄(aq)

no of moles of Na2B4O7(s) = weight / molar mass = 5.18 / 201.22 = 0.02574 moles

no ofmoles of H2SO4 = 5.18 / 98.1 = 0.053 moles

no of moles ofH2O = 5.18 / 18 = 0.29 moles

1 mole Na2B4O7 reacts to produce 1 mole Na2SO4
So moles Na2SO4 possible = moles Na2B4O7 = 0.02574 mol

1 mole H2SO4 reacts to produce 1 mole Na2SO4
So moles Na2SO4 possible = 1 mole H2SO4 = 0.053 mol

5 moles H2O react to from 1 mole Na2SO4
Therefore moles Na2SO4 = 1/5 x moles H2O = 0.29 / 5 = 0.058 mol

The limiitng reagent is Na2B4O7, the amount provided will produce the least amount of product

Now, 1 mole Na2B4O7 reacts with 5 moles H2O

So moles H2O needed to fully react all the Na2B4O7 = 5 x moles Na2B4O7

= 5 x 0.0574 mol

= 0.1287 moles H2O

So to fully react all the limiting reagent uses up 0.1287 moles of H2O

moles H2O left over = moles H2O provided - moles used

= 0.29 mol - 0.1287 mol

= 0.1613 moles remaining

mass = molar mass x moles

mass H2O = 18.016 g/mol x 0.1613 mol

= 2.906 grams