When reviewing the procedure the student found that 4.912 x 10-1 M barium nitrate solution had been used instead of that in the following procedure:
A student synthesized 6.895 g of barium iodate monohydrate, Ba(IO3)2*H2O by adding 30.00 mL of 5.912 x 10-1 M barium nitrate, Ba(NO3)2, to 50.00 mL of 9.004 x 10-1 M sodium Iodate, NaIO3
Calculate the theoretical yield of barium iodate monohydrate using 30.00 mL of 4.912 x 10-1 M barium nitrate solution
Ba(NO3)2 + 2NaIO3 + H2O ==> Ba(IO3)2*H2O + 2NaNO3
Ba(NO3)2 is the limiting reagent with either number.
mols Ba(NO3)2 = M x L = 0.4912 x 0.030 = 0.014736 = mols Ba(IO3)2.H2O
g Ba(IO3)2 = mols x molar mass = theoretical yield (TY). Actual yield (AY) = 6.895g in the problem.
% yield (but not asked for above) = (AY/TY)*100 = ?
To calculate the theoretical yield of barium iodate monohydrate using 30.00 mL of 4.912 x 10^-1 M barium nitrate solution, we need to follow these steps:
1. Convert the concentration of barium nitrate solution from molarity (M) to moles per liter (mol/L):
Concentration = 4.912 x 10^-1 M = 4.912 x 10^-1 mol/L
Note: Molarity (M) represents moles of solute per liter of solution.
2. Calculate the number of moles of barium nitrate in 30.00 mL of the solution:
Volume = 30.00 mL = 30.00 cm^3 = 30.00 x 10^-3 L
Moles of Ba(NO3)2 = Concentration x Volume
= (4.912 x 10^-1 mol/L) x (30.00 x 10^-3 L)
3. Use the balanced chemical equation for the reaction between barium nitrate and sodium iodate to determine the molar ratio between barium nitrate and barium iodate monohydrate:
Ba(NO3)2 + NaIO3 → Ba(IO3)2 + 2 NaNO3
The coefficient of Ba(NO3)2 in the balanced equation is 1, which means that 1 mole of Ba(NO3)2 reacts to produce 1 mole of Ba(IO3)2.
4. Calculate the number of moles of Ba(IO3)2 that can be produced based on the moles of Ba(NO3)2:
Moles of Ba(IO3)2 = Moles of Ba(NO3)2
5. Determine the molar mass of Ba(IO3)2*H2O (barium iodate monohydrate):
Molar mass = (1 mole of Ba x atomic mass of Ba) + (2 moles of I x atomic mass of I) + (6 moles of O x atomic mass of O) + (2 moles of H x atomic mass of H) + (1 mole of O x atomic mass of O)
= (1 x 137.33 g/mol) + (2 x 126.90 g/mol) + (6 x 16.00 g/mol) + (2 x 1.01 g/mol) + (1 x 16.00 g/mol)
6. Calculate the mass of barium iodate monohydrate that can be produced using the moles and molar mass:
Mass = Moles of Ba(IO3)2 * Molar mass
This will give you the theoretical yield of barium iodate monohydrate using 30.00 mL of 4.912 x 10^-1 M barium nitrate solution.