Question

If 25.0 grams of Sb2S3 reacts with an excess of hydrochloric acid, how many grams of H2S are formed? What volume does the H2S formed occupy under conditions of STP?

Answer 1

Answer: a) 7.57 g

b) 4.97 L

Step-by-step explanation:

To calculate the moles :

`\text{Moles of solute}=\frac{\text{given mass}}{\text{Molar Mass}}`

`\text{Moles of} Sb_2S_3=(25.0g)/(339.7g/mol)=0.074moles`

`Sb_2S_3+6HCl\rightarrow 3H_2S+2SbCl_3`

As
`HCl` is the excess reagent,
`Sb_2S_3` is the limiting reagent and it limits the formation of product.

According to stoichiometry :

1 mole of
`Sb_2S_3` form = 3 moles of
`H_2S`

Thus 0.074 moles of
`Sb_2S_3` will form =
`(3)/(1)* 0.074=0.222moles` of
`H_2S`

Mass of
`H_2S=moles* {\text {Molar mass}}=0.222moles* 34.1g/mol=7.57g`

According to ideal gas equation:

`PV=nRT`

P = pressure of gas = 1 atm (at STP)

V = Volume of gas = ?

n = number of moles = 0.222

R = gas constant =
`0.0821Latm/Kmol`

T =temperature =
`273K`

`V=(nRT)/(P)`

`V=(0.222* 0.0821L atm/K mol* 273K)/(1)=4.97L`

Thus the volume occupied by
`H_2S` is 4.97 L