Question

Determine the percent yield for an acid/base reaction that involves mixing 500. mL of 0.100 molar nitric acid with 500. mL of 0.0500 molar barium hydroxide and a 0.00200 molar solution of barium nitrate is produced.

Answer 1

Solution :-

Balanced reaction equation is as follows

Ba(OH)2 + 2 HNO3 ----- > Ba(NO3)2 + 2 H2O

now lets calculate the moles of the HNO3 and Ba(OH)2 using the molarities and volumes

moles = molarity * volume in liter

moles of HNO3 = 0.100 mol per L * 0.500 L = 0.0500 mol

moles of Ba(OH)2 = 0.0500 mol per L * 0.500 L = 0.025 mol

mole ratio of the reactant is 1 :2 for the Ba(OH)2 to HNO3

so the moles of the both reactant are exactly the same that needed to react with one another

now lets calculate the moles of the product.

0.025 mol Ba(OH)2 * 1 mol Ba(NO3)2 / 1 mol Ba(OH)2 = 0.025 mol Ba(NO3)2

after the mixing of solutions the final volume is 500 ml + 500 ml = 1000 ml = 1.00 L

so lets calculate the theoreticall molarity of the Ba(NO3)2

molarity = moles / volume in liter

molarity of the Ba(NO3)2 = 0.025 mol / 1.00 L = 0.0250 M

now lets calculate the peercent yield

% yield = (actual molarity / theoretical molarity) *100 %

           = (0.002/0.025)*100%

           = 8.00 %

Therefore the percent yield = 8.00 %