Ex 3:24.0 g of Iodine react with 0.20 g of hydrogen to produce 11.8 g ofhydrogen iodide. What is the percent yield of this reaction?

To calculate the percent yield of a reaction, we first need to know the theoretical yield. The theoretical yield is the maximum amount of product that can be produced from a given amount of reactant.

Step 1: Calculate the moles of reactants The molar mass of iodine (I2) is approximately 253.8 g/mol and the molar mass of hydrogen (H2) is approximately 2.02 g/mol.

Moles of I2 = 24.0 g / 253.8 g/mol = 0.0945 mol Moles of H2 = 0.20 g / 2.02 g/mol = 0.0990 mol

Step 2: Determine the limiting reactant The reaction between iodine and hydrogen to produce hydrogen iodide is as follows:

I2 + H2 -> 2HI

From the balanced equation, we can see that one mole of I2 reacts with one mole of H2 to produce two moles of HI. Therefore, the limiting reactant is I2 because we have less moles of I2 than H2.

Step 3: Calculate the theoretical yield The theoretical yield is the amount of product that would be produced if all of the limiting reactant reacted. From the balanced equation, we can see that one mole of I2 produces two moles of HI. Therefore, the theoretical yield is:

0.0945 mol I2 x 2 mol HI/1 mol I2 = 0.189 mol HI

The molar mass of HI is approximately 127.9 g/mol, so the theoretical yield in grams is:

0.189 mol HI x 127.9 g/mol = 24.2 g

Step 4: Calculate the percent yield The percent yield is the actual yield (the amount of product actually produced) divided by the theoretical yield, multiplied by 100%. The actual yield is given as 11.8 g. Therefore, the percent yield is:

(11.8 g / 24.2 g) x 100% = 48.8%

So, the percent yield of the reaction is 48.8%.