For the following reaction, 0.372 grams of hydrogen gas are allowed to react with 50.1 grams of iodine.
hydrogen(g) + iodine(s) hydrogen iodide(g)
1. What is the maximum mass of hydrogen iodide that can be formed?
grams
2. What is the FORMULA for the limiting reagent?
3. What mass of the excess reagent remains after the reaction is complete? grams
H2(g) + I2(s) ==> 2HI(g)
mols H2 = g/molar mass = 0.372/2 = 0.186
mols I2 = 50.1/127*2 = 0.197
If H2 is the limiting reagent(LR) it will require 0.186 mols of I2. You have that much I2; therefore, H2 is the LR and I2 is the ER (excess reagent).
HI formed: mols H2 x (2 mols HI/1 mol H2) = 0.186 mols H2 x 2 = 0.372 g HI. That is the maximum amount that can be formed. Convert to grams by grams = mols x molar mass = ?
ER remaining:
0.197 = mols I2 to start
-0.186 = mols I2 used
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xxxx you do it.= mols I2 remaining.
grams = mols x molar mass = ?
Post your work if you get stuck.