Question

How many grams of HI can be made from 6 grams of $H_2$ and 800 grams of $I_2$ in the following reaction: $H_2+I_2\to 2HI$?

• A. 800 grams of HI can be made with 38 grams of excess iodine
• B. 768 grams of HI can be made with 6 grams of excess hydrogen
• C. 768 grams of HI can be made with 38 grams of excess iodine
• D. 2286 grams of HI can be made with no excess reactants
• E. 806 grams of HI can be made with no excess reactants

Answer 1

The correct option is C 768 grams of HI can be made with 38 grams of excess iodine

The molar masses of $H_2$, $I_2$ and $HI$ are 2 g/mol, 254 g/mol and 128 g/mol respectively.
The number of moles is the ratio of mass to molar mass.

The number of moles of $H_2$ $=\frac{6g}{2g/mol}=3$ moles

The number of moles of $I_2$ $=\frac{800g}{254g/mol}=3.15$ mole.
1 mole of $H_2$ will react with 1 mole of $I_2$.
Hence, 3 moles of $H_2$ will react with 3 moles of $I_2$.
Hence, $H_2$ is the limiting reagent and iodine is the excess reagent.
$3.15-3=0.15$ moles of $I_2$ is in excess
$0.15\times 254=38$ grams of $I_2$ is in excess.
3 moles of $H_2$ will give
$3\times 2=6$ moles of $HI$
The mass of HI obtained is $6\times 128=768$ grams.