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
whats the percent yield?
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To determine the percent yield, we need to compare the actual yield to the theoretical yield. The actual yield is the amount of barium iodate monohydrate that was synthesized, which is given as 6.895 g. The theoretical yield is the maximum amount of barium iodate monohydrate that could be obtained based on stoichiometry and the limiting reactant.
First, we need to determine the limiting reactant - the reactant that is consumed completely and determines the amount of product that can be formed. To find the limiting reactant, we compare the number of moles of each reactant.
The molar mass of barium nitrate, Ba(NO3)2, is calculated as follows:
Ba(NO3)2 = (1 Ba) + (2 N) + (6 O) = 137.33 g/mol.
To find the number of moles of barium nitrate, we use the molarity and volume:
moles of Ba(NO3)2 = (0.5 L) x (5.912 x 10^-1 M) = 2.956 x 10^-1 mol.
The molar mass of sodium iodate, NaIO3, can be calculated as follows:
NaIO3 = (1 Na) + (1 I) + (3 O) = 197.89 g/mol.
To find the number of moles of sodium iodate, we use the molarity and volume:
moles of NaIO3 = (0.05 L) x (9.004 x 10^-1 M) = 4.502 x 10^-2 mol.
Next, we need to find the mole ratio between the reactants and the product. From the balanced equation:
Ba(NO3)2 + 2NaIO3 -> Ba(IO3)2 + 2NaNO3.
The ratio of barium nitrate to barium iodate monohydrate is 1:1, meaning that 1 mole of barium nitrate reacts with 1 mole of barium iodate monohydrate.
Since the moles of barium nitrate and sodium iodate are both larger than the mole ratio of barium nitrate to barium iodate monohydrate, we can conclude that the limiting reactant is the one with the smallest number of moles, which is sodium iodate.
Now we can calculate the theoretical yield using the stoichiometry. The molar mass of barium iodate monohydrate, Ba(IO3)2*H2O, is calculated as follows:
Ba(IO3)2*H2O = (1 Ba) + (2 I) + (2 H2O) + (6 O) = 433.20 g/mol.
The moles of barium iodate monohydrate can be calculated from the moles of sodium iodate:
moles of Ba(IO3)2*H2O = (4.502 x 10^-2 mol) x (1 mol Ba(IO3)2*H2O / 2 mol NaIO3) = 2.251 x 10^-2 mol.
Now we can find the theoretical yield in grams:
theoretical yield = (2.251 x 10^-2 mol) x (433.20 g/mol) = 9.758 g.
Finally, we can calculate the percent yield:
percent yield = (actual yield / theoretical yield) x 100%
percent yield = (6.895 g / 9.758 g) x 100%
percent yield = 70.7%
Therefore, the percent yield of the reaction is 70.7%.