Question

Hydrogen iodide, HI, is formed in an equilibrium reaction when gaseous hydrogen and iodine gas are heated together. If 20.0 g of hydrogen and 20.0 g of iodine are heated, forming 10.0 g of hydrogen iodide, what mass of hydrogen remains unreacted? A. 10.0 g hydrogen remains B. 10.9 g hydrogen remains C. 15.0 g hydrogen remains D. 19.9 g hydrogen remains.

Answer 1

Answer: D. 19.9 g hydrogen remains.

Step-by-step explanation:

To calculate the moles, we use the equation:

`\text{Number of moles}=\frac{\text{Given mass}}{\text {Molar mass}}`

a) moles of
`H_2`

`\text{Number of moles}=(20.0g)/(2g/mol)=10.0moles`

b) moles of
`I_2`

`\text{Number of moles}=(20.0g)/(254g/mol)=0.0787moles`

`H_2(g)+I_2(g)\rightarrow 2HI(g)`

According to stoichiometry :

1 mole of
`I_2` require 1 mole of
`H_2`

Thus 0.0787 moles of
`l_2` require=
`(1)/(1)* 0.0787=0.0787moles` of
`H_2`

Thus
`l_2` is the limiting reagent as it limits the formation of product and
`H_2` acts as the excess reagent. (10.0-0.0787)= 9.92 moles of
`H_2`are left unreacted.

Mass of
`H_2=moles* {\text {Molar mass}}=9.92moles* 2.01g/mol=19.9g`

Thus 19.9 g of
`H_2` remains unreacted.