For the following reaction, 6.79 grams of hydrogen gas are mixed with excess iodine . The reaction yields 775 grams of hydrogen iodide . hydrogen ( g ) iodine ( s ) hydrogen iodide ( g ) What is the theoretical yield of hydrogen iodide

Explanation:

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In this case, for the reaction:

As 6.79 g of hydrogen (molar mass = 2.02 g/mol) react in excess iodine, we can compute the theoretical yield of hydrogen iodide (molar mass = 127.91 g/mol)  via their 1:2 mole ratio shown on the chemical reaction:

Next, we compute the percent yield by divinding the actual yield (775 g) by the theoretical yield (860 g):

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you need to take a look at the periodic system in order to sort these elements into pairs that would most likely exhibit similar chemical properties. first, you need to find elements that are found in the same group (groups are vertical in the periodic system), and then to see whether they have the same valence number. if they meet these criteria, they will have similar properties.

the first pair: ne (neon) and kr (krypton), are noble gases found in group 8

the second pair: li (lithium) and k (potassium), found in group 1

the third pair: s (sulfur) and te (tellurium), found in group 6

i dont know the answe