Question

1.Following the general procedure described in this experiment , a student synthesized 6.895 g of barium idoate monhydrate, Ba(IO3)2 . H2), by adding 30.00 mL of 5.912x10^(-1) M barium nitrate, Ba(NO3)2, to 50.00 mL of 9.004x10^(-1) M sodium idoate, NaIO3.

(A) Write the chemical equation for the reaction of solutions of barium nitrate and sodium iodate.

(B) Calculate the precent yield of barium iodate monohydrate the student obtaned in this experiment.

When reviewing the procedure and calculations, the student discovered that a 4.912x10^(-1) M barium nitrate solution had been used instead of a 5.912x10^(-1) M solution. (C) Calculate the precent yield of barium iodate monhydrate using 30.00 mL of 4.912x10^(-1) M barium nitrate solution and 50.00 mL of 9.004x10^(-1) M sodium iodate solution.

(4) Calculate the precent error in the precent yield calculated in (2) compared with the correct precent yield calculated in (3)

2. Assume that, in the experment described in question (1), 125 mL of 25 C distilled water was used to wash and transfer the precipitate, rather than 20 mL of chilled distilled water (at4 C). The solubility of barium iodate monohydrate in 25 C water is 0.028 g per 100 mL of water; in 4 C water, it is 0.010 g per 100 mL of water.

(1) What mass of product would you expect to isolate?

(2) Calculate the precent error as a result of using 125 mL of 25 C water, compared with the correct yield using 20 mL of 4 C water

Answer 1

a) the chemical reaction is as follows,

Ba(NO3)2 + 2NaIO3 + H2O ==> Ba(IO3)2.H2O + 2NaNO3

b) moles of Ba(NO3)2 = 5.912x10^(-1) M * 0.03 = 0.0177 moles

moles of NaIO3 = 9.004x10^(-1) M * 0.05 = 0.4502 moles

hence, Ba(NO3)2 is the limiting reactant.

from the balanced reaction the ratio of Ba(NO3)2 : Ba(IO3)2.H2O = 1:1

hence, moles of Ba(IO3)2.H2O = 0.0177 moles

theorectical yield of Ba(IO3)2.H2O =  0.0177 * 507.2911 g/mol = 8.997g

% yied = (6.895 g / 8.997) *100 = 76.63%

C)

moles of Ba(NO3)2 = 4.912x10^(-1) M * 0.03 = 0.0147 moles

moles of NaIO3 = 9.004x10^(-1) M * 0.05 = 0.4502 moles

hence, Ba(NO3)2 is the limiting reactant.

from the balanced reaction the ratio of Ba(NO3)2 : Ba(IO3)2.H2O = 1:1

hence, moles of Ba(IO3)2.H2O = 0.0147 moles

theorectical yield of Ba(IO3)2.H2O =  0.0147 * 507.2911 g/mol = 7.4571g

% yied = (6.895 g / 7.4571) *100 = 92.46%

d) % error =( difference / actual ) *100

= 17.11%